• 2022天梯赛练习集（2022.9-2022.10）

    使用函数判断完全平方数 

    没有加（int）过不了 

    1. int IsSquare(int n){
    2. 	if((int)sqrt(n) * sqrt(n) != n) return 0;
    3. 	else return 1;
    4. }

    使用函数求余弦函数的近似值 

    1. double funcos(double e, double x){
    2.     double sum = 1, item = 1;
    3.     for(int i = 0; fabs(item) >= e; i += 2){
    4.         item = (-1) * item * x * x / ((i + 1) * (i + 2));
    5.         sum += item;
    6.     }
    7.     return sum;
    8. }

    6-13 使用函数输出水仙花数

    1. int narcissistic(int number){
    2. 	int n = number;
    3.     int item = 0, sum = 0, cnt = 0;
    4. 	while(n){
    5. 		n /= 10;
    6.         cnt ++;
    7. 	}
    8. 	item = number;
    9.     while(item){
    10.         int lastnumber = item % 10;
    11.         sum += pow(lastnumber, cnt);
    12.         item /= 10;
    13.     }
    14.     if(sum == number) return 1;
    15. 	return 0;
    16. 	
    17. }
    18. void PrintN(int m, int n){
    19. 	for(int i = m + 1; i < n; i ++ ){
    20. 		if(narcissistic(i)) printf("%d\n", i);
    21. 	}
    22. }

    6-18 使用函数输出指定范围内的完数 

    在一些高级语言当中，为了能够完成更好的逻辑判断，因此就有了bool类型，bool类型的变量值只有true和false两种。

    而在C语言中，一般认为0为假，非0为真。
    这是因为c99之前，c90是没有bool类型的的。但是c99引入了_Bool类型（_Bool就是一个类型，不过在新增头文件stdbool.h中，被重新用宏写成了 bool，为了保证C/C++兼容性）。
    目前为止大部分C语言书籍采用的标准还是c90标准，因此我们很少用bool类型。
     

    factorsum函数中如果是

    1. int factorsum( int number ){
    2.     int sum = 1;
    3.     for(int i = 2; i < number; i ++ ){
    4.         if(number % i == 0) sum += i;
    5.     }
    6.     return sum;
    7. }

    过不了 最大范围 的测试点

    1. int factorsum( int number ){
    2.     int sum = 0;
    3.     for(int i = 1; i < number; i ++ ){
    4.         if(number % i == 0) sum += i;
    5.     }
    6.     return sum;
    7. }
    8. void PrintPN( int m, int n ){
    9.     int flag = 0;
    10.     for(int i = m; i <= n; i ++ ){
    11.         if(factorsum(i) == i){
    12.             flag = 1;
    13.             printf("%d = 1", i);
    14.             for(int j = 2; j < i; j ++ ){
    15.                 if(i % j == 0) printf(" + %d", j);
    16.             }
    17.             printf("\n");
    18.         }
    19.     }
    20.     if(!flag) printf("No perfect number");
    21. }

    6-19 使用函数输出指定范围内的Fibonacci数

    一开始的代码一直出现 [PTA报错]warning: ignoring return value of ‘scanf’, declared with attribute warn_unused_result 的问题

    下面的ac代码反向使用cnt，和我平时的用法感觉妙了点（？

    1. int fib( int n ){
    2.     if(n == 1 || n == 2) return 1;
    3.     return fib(n - 1) + fib(n - 2);
    4. }
    5. void PrintFN( int m, int n ){
    6.     int cnt = 1, flag = 0;
    7.     for(int i = 1; fib(i) <= n; i ++ ){
    8.         if(fib(i) >= m && fib(i) <= n){
    9.             if(cnt){
    10.                 printf("%d", fib(i));
    11.                 cnt = 0;
    12.             }
    13.             else printf(" %d", fib(i));
    14.             flag = 1;
    15.         }
    16.     }
    17.     if(!flag) printf("No Fibonacci number");
    18. }

    6-30 删除字符

    1. void delchar( char *str, char c ){
    2.     int j = 0;
    3.     for(int i = 0; *(str + i) !='\0'; i ++ ){
    4.         if(*(str + i) != c){
    5.             *(str + j) = *(str + i);
    6.             j ++;
    7.         }
    8.     }
    9.     *(str + j) = '\0';
    10. }

    6-31 字符串的连接 

    1. char *str_cat( char *s, char *t ){
    2.     int len_s = strlen(s);
    3.     int len_t = strlen(t);
    4.     for(int i = 0; i < len_t; i ++ ){
    5.         *(s + len_s + i) = *(t + i);
    6.     }
    7.     return &s[0];
    8. }

    6-46 指定位置输出字符串 

    1. char *match( char *s, char ch1, char ch2 ){
    2.     char *p;
    3.     int i;
    4.     for(i = 0; i < strlen(s); i ++ ){
    5.         if(s[i] == ch1) break;
    6.     }
    7.     p = &s[i];
    8.     for(int j = i; j < strlen(s); j ++ ){
    9.         printf("%c", s[j]);
    10.         if(s[j] == ch2) break;
    11.     }
    12.     printf("\n");
    13.     return p;
    14. }

    6-47 查找子串 

    1. char *search( char *s, char *t ){
    2.     int flag = 0;
    3.     for(int i = 0; s[i] != '\0'; i ++ ){
    4.         for(int j = 0; t[j] != '\0'; j ++ ){
    5.             flag = 0;
    6.             if(s[i + j] != t[j]) break;
    7.             else flag = 1;
    8.         }
    9.         if(flag) return &s[i];
    10.     }
    11.     return NULL;
    12. }

    6-49 建立学生信息链表

    1. void input(){
    2.     struct stud_node *p;
    3.     p = (struct stud_node *)malloc(sizeof(struct stud_node));
    4.     scanf("%d", &p->num);
    5.     while(p->num != 0){
    6.         scanf("%s %d", p->name, &p->score);
    7.         if(head == NULL){
    8.             head = p;
    9.             head->next = NULL;
    10.         }
    11.         if(tail != NULL) tail->next = p;
    12.         tail = p;
    13.         tail->next = NULL;
    14.         p = (struct stud_node *)malloc(sizeof(struct stud_node));
    15.         scanf("%d", &p->num);
    16.     }
    17. }

    6-50 学生成绩链表处理 

    1. struct stud_node *createlist(){
    2.     struct stud_node *head, *tail, *s;
    3.     head = (struct stud_node *)malloc(sizeof(struct stud_node));
    4.     head->next = NULL;
    5.     tail = head;
    6.     s = (struct stud_node *)malloc(sizeof(struct stud_node));
    7.     while(1){
    8. 		s=(struct stud_node*)malloc(sizeof(struct stud_node));
    9. 		scanf("%d",&s->num);
    10. 		if(s->num==0)break;
    11. 		scanf("%s",s->name);
    12. 		scanf("%d",&s->score);
    13. 		s->next=NULL;
    14. 		tail->next=s;
    15. 		tail=s;
    16. 	}
    17.     return head;
    18. }
    19. struct stud_node *deletelist( struct stud_node *head, int min_score ){
    20.     struct stud_node *p, *n;
    21.     p = head->next;
    22.     while(p != NULL){
    23.         if(p->score < min_score){
    24.             n = head;
    25.             while(n->next != p) n = n->next;
    26.             n->next = p->next;
    27.             free(p);
    28.         }
    29.         p = p->next;
    30.     }
    31.     return head->next;
    32. }

    6-51 逆序数据建立链表 

    1. struct ListNode *createlist(){
    2.     struct ListNode *head, *tail, *p;
    3.     head = tail = NULL;
    4.     int num;
    5.     while(scanf("%d", &num) && num != -1){
    6.         p = (struct ListNode *)malloc(sizeof(struct ListNode));
    7.         p->data = num;
    8.         p->next = head;
    9.         head = p;
    10.     }
    11.     return head;
    12. }

    6-52 链表拼接 

    1. struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2){
    2.     struct ListNode *head;
    3.     head = (struct ListNode *)malloc(sizeof(struct ListNode));
    4.     struct ListNode *p = head, *p1 = list1, *p2 = list2;
    5.     while(p1 && p2){
    6.         if(p1->data < p2->data){
    7.             p->next = p1;
    8.             p1 = p1->next;
    9.             p = p->next;
    10.         }
    11.         else{
    12.             p->next = p2;
    13.             p2 = p2->next;
    14.             p = p->next;
    15.         }
    16.     }
    17.     p->next = p1 ? p1 : p2;
    18.     return head->next;
    19. }

    6-53 奇数值结点链表 

    1. struct ListNode *readlist(){
    2.     struct ListNode *head = NULL;
    3.     struct ListNode *current, *prev;
    4.     int n;
    5.     while(scanf("%d", &n) == 1 && n != -1){
    6.         current = (struct ListNode *)malloc(sizeof(struct ListNode));
    7.         if(head == NULL) head = current;
    8.         else prev->next = current;
    9.         current->next = NULL;
    10.         current->data = n;
    11.         prev = current;
    12.     }
    13.     return head;
    14. }
    15. struct ListNode *getodd( struct ListNode **L ){
    16.     struct ListNode *t = *L;
    17.     struct ListNode *head_even = NULL, *head_odd = NULL;
    18.     struct ListNode *current_even, *current_odd, *prev_even, *prev_odd;
    19.     while(t){
    20.         if((t->data % 2) == 0){
    21.             current_even = (struct ListNode *)malloc(sizeof(struct ListNode));
    22.             if(head_even == NULL) head_even = current_even;
    23.             else prev_even->next = current_even;
    24.             current_even->next = NULL;
    25.             current_even->data = t->data;
    26.             prev_even = current_even;
    27.         }
    28.         else{
    29.             current_odd = (struct ListNode *)malloc(sizeof(struct ListNode));
    30.             if(head_odd == NULL) head_odd = current_odd;
    31.             else prev_odd->next = current_odd;
    32.             current_odd->next = NULL;
    33.             current_odd->data = t->data;
    34.             prev_odd = current_odd;
    35.         }
    36.         t = t->next;
    37.     }
    38.     *L = head_even;
    39.     return head_odd;
    40. }

    6-54 单链表结点删除 

    hhhhh,仿照上面的自己敲出来了红红火火恍恍惚惚哈哈哈（已疯 

    1. struct ListNode *readlist(){
    2.     struct ListNode *head = NULL;
    3.     struct ListNode *prev, *current;
    4.     int n;
    5.     while(scanf("%d", &n) == 1 && n != -1){
    6.         current = (struct ListNode *)malloc(sizeof(struct ListNode));
    7.         if(head == NULL) head = current;
    8.         else prev->next = current;
    9.         current->next = NULL;
    10.         current->data = n;
    11.         prev = current;
    12.     }
    13.     return head;
    14. }
    15. struct ListNode *deletem( struct ListNode *L, int m ){
    16.     struct ListNode *t = L;
    17.     struct ListNode *head = NULL;
    18.     struct ListNode *current, *prev;
    19.     while(t){
    20.         if(t->data != m){
    21.             current = (struct ListNode *)malloc(sizeof(struct ListNode));
    22.             if(head == NULL) head = current;
    23.             else prev->next = current;
    24.             current->next = NULL;
    25.             current->data = t->data;
    26.             prev = current;
    27.         }
    28.         t = t->next;
    29.     }
    30.     return head;
    31. }

    6-55 链表逆置 

    1. struct ListNode *reverse( struct ListNode *head ){
    2.     struct ListNode *Head, *p, *q;
    3.     p = head;
    4.     Head = NULL;
    5.     while(p){
    6.         q = p->next;
    7.         p->next = Head;
    8.         Head = p;
    9.         p = q;
    10.     }
    11.     return Head;
    12. }

    6-56 统计专业人数 

    1. int countcs( struct ListNode *head ){
    2.     int cnt = 0;
    3.     if(head == NULL) return cnt;
    4.     while(head){
    5.         char code[8];
    6.         strcpy(code, head->code);
    7.         if(code[1] == '0' && code[2] == '2') cnt ++;
    8.         head = head->next;
    9.     }
    10.     return cnt;
    11. }

     6-57 删除单链表偶数节点

    这道题就是前面单链表结点删除直接复制粘贴改几个就ac了（ 

    1. struct ListNode *createlist(){
    2.     struct ListNode *head = NULL;
    3.     struct ListNode *prev, *current;
    4.     int n;
    5.     while(scanf("%d", &n) == 1 && n != -1){
    6.         current = (struct ListNode *)malloc(sizeof(struct ListNode));
    7.         if(head == NULL) head = current;
    8.         else prev->next = current;
    9.         current->next = NULL;
    10.         current->data = n;
    11.         prev = current;
    12.     }
    13.     return head;
    14. }
    15. struct ListNode *deleteeven( struct ListNode *head ){
    16.     struct ListNode *t = head;
    17.     struct ListNode *Head = NULL;
    18.     struct ListNode *current, *prev;
    19.     while(t){
    20.         if(t->data % 2 != 0){
    21.             current = (struct ListNode *)malloc(sizeof(struct ListNode));
    22.             if(Head == NULL) Head = current;
    23.             else prev->next = current;
    24.             current->next = NULL;
    25.             current->data = t->data;
    26.             prev = current;
    27.         }
    28.         t = t->next;
    29.     }
    30.     return Head;
    31.     
    32. }

     7-45 高速公路超速处罚

    打%号 %%

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 40;
    6. int main(){
    7. 	int v, lv;
    8. 	cin >> v >> lv;
    9. 	double x = 100.0 * (v - lv) / lv;
    10. 	if(x < 10) cout << "OK";
    11. 	else if(x < 50) printf("Exceed %.lf%%. Ticket 200\n", x);
    12. 	else printf("Exceed %.lf%%. License Revoked\n", x);
    13. 	return 0;
    14. }

     7-54 猜数字游戏

    if else 过不了，要用if if if，因为后面还有一个循环外的条件不能都过？？就是感觉这题还是不太会wwwww

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 40;
    6. int main(){
    7. 	int sn, n;
    8. 	cin >> sn >> n;
    9. 	int cnt = 0;
    10. 	for(int i = 1; i <= n; i ++ ){
    11. 		int a;
    12. 		cin >> a;
    13. 		cnt ++ ;
    14. 		
    15. 		if(a == sn && cnt == 1){
    16. 			cout << "Bingo!" << endl;
    17. 			break;
    18. 		}
    19. 		else if(a == sn && cnt <= 3 && cnt >= 1){
    20. 			cout << "Lucky you!" << endl;
    21. 			break;
    22. 		}
    23. 		else{
    24. 			if(a < 0){
    25. 				cout << "Game over" << endl;
    26. 				break;
    27. 			}
    28. 			if(a > sn){
    29. 				cout << "Too big" << endl;
    30. 			}
    31. 			if(a < sn){
    32. 				cout << "Too small" << endl;
    33. 			}
    34.             if(a == sn){
    35.                 cout << "Good Guess!" << endl;
    36.                 break;
    37.             }
    38. 		}
    39. 		if(a != sn && cnt == n){
    40. 			cout << "Game over" << endl;
    41. 		}
    42. 	}
    43. 	return 0;
    44. }

    7-56 高空坠球

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 40;
    6.  
    7. int main(){
    8. 	double H, n;
    9. 	cin >> H >> n;
    10. 	if(n == 0){
    11. 		printf("0.0 0.0");
    12.         return 0;
    13. 	} 
    14. 	double sum = 0, h = H;
    15. 	for(int i = 1; i <= n; i ++ ){
    16. 		sum += h;
    17. 		h /= 2;
    18. 		sum += h;
    19. 	}
    20. 	printf("%.1lf %.1lf", sum - h, h);
    21. 	return 0;
    22. }

    7-57 求分数序列前N项和 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 40;
    6.  
    7. int main(){
    8. 	int n;
    9. 	cin >> n;
    10. 	double sum  = 0;
    11. 	double z = 2.0, m = 1.0;
    12. 	for(int i = 1; i <= n; i ++ ){
    13. 		sum += z / m;
    14. 		double t = z; 
    15. 		z += m;
    16. 		m = t;
    17. 	}
    18. 	printf("%.2lf", sum);
    19. 	return 0;
    20. }

    7-58 求e的近似值 

    自定的fun函数int型过不了，要double

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 40;
    6. double fun(int n){
    7. 	double ans = 1;
    8. 	for(int i = 1; i <= n; i ++ ){
    9. 		ans *= i;
    10. 	}
    11. 	return ans;
    12. }
    13. int main(){
    14. 	int n;
    15. 	cin >> n;
    16. 	double sum = 1; 
    17. 	for(int i = 1; i <= n; i ++ ){
    18. 		sum += 1.0 / fun(i);
    19. 	}
    20. 	printf("%.8lf", sum);
    21. 	return 0;
    22. }

    7-61 求幂级数展开的部分和 

    a啊啊啊啊啊·，还是没做出来，超时了一个点QAQ 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. //const int N = 40;
    6. int main(){
    7. 	double x;
    8. 	cin >> x;
    9. 	double sum = 1.0, ans = 1.0;
    10. 	for(int i = 1;; i ++ ){
    11. 		ans = (ans * x) / i;
    12.         sum += ans;
    13.         if(ans < 1e-5) break;
    14. 	}
    15. 	printf("%.4lf", sum);
    16. 	return 0;
    17. }

    7-71 交换最小值和最大值 

    交换位置的时候amax要在amin前面，不然 最小值出现最后 的测试点不能过

    分析一下，如果最小值在最后，最大值在最前面的情况下：

    eg:8 2 5 4 1

    ac的代码来分析，先找到的是和amax相等的a[0]， 交换后就变成了 1 2 5 4 8

    继续循环，然后找到了和amin相等的a[0]， 交换后变成了 1 2 5 4 8

    拿 最小值出现最后 的测试点不能过的代码来说：

    eg:8 2 5 4 1

    先找到的是和amax相等的a[0]， 交换后变成了1 2 5 4 8

    但是这时候只能继续循环开始 i = 1 了，可是此时最小值在 i = 0 , 所以无法交换最小值到最后的位置 

    1. //最小值出现在最后的测试点过不了
    2. for(int i = 0; i < n; i ++ ){
    3. 		if(amin == a[i]){
    4. 			a[i] = a[0] ^ a[i];
    5. 			a[0] = a[0] ^ a[i];
    6. 			a[i] = a[0] ^ a[i];
    7. 		}
    8. 		if(amax == a[i]){
    9. 			a[n - 1] = a[n - 1] ^ a[i];
    10. 			a[i] = a[n - 1] ^ a[i];
    11. 			a[n - 1] = a[n - 1] ^ a[i];
    12. 		}
    13. 	}

     AC代码

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 10;
    6. int a[N];
    7. int main(){
    8. 	int n;
    9. 	cin >> n;
    10. 	for(int i = 0; i < n; i ++ ) cin >> a[i];
    11. 	int amin = a[0], amax = a[0];
    12. 	for(int i = 1; i < n; i ++ ){
    13. 		amin = min(amin, a[i]);
    14. 		amax = max(amax, a[i]);
    15. 	}
    16. 	for(int i = 0; i < n; i ++ ){
    17.         if(amax == a[i]){
    18. 			int t = a[n - 1];
    19.             a[n - 1] = amax;
    20.             a[i] = t;
    21. 		}
    22. 		if(amin == a[i]){
    23. 			int t = a[0];
    24.             a[0] = amin;
    25.             a[i] = t;
    26. 		}
    27. 	}
    28. 	for(int i = 0; i < n; i ++ ) cout << a[i] << " ";
    29. 	return 0;
    30. }

    7-75 找出不是两个数组共有的元素 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 25;
    6. int a[N], b[N], c[N];
    7. int main(){
    8.     int n, m, i, j;
    9.     int k = 0;
    10.     cin >> n;
    11.     for(i = 0; i < n; i ++ ) cin >> a[i];
    12.     cin >> m;
    13.     for(j = 0; j < m; j ++ ) cin >> b[j];
    14.     for(i = 0; i < n; i ++ ){
    15.     	for(j = 0; j < m; j ++ ){
    16.     		if(a[i] == b[j]) break;
    17. 		}
    18. 		if(j == m) c[k ++ ] = a[i];
    19. 	}
    20. 	for(i = 0; i < m; i ++ ){
    21. 		for(j = 0; j < n;  j ++ ){
    22. 			if(a[j] == b[i]) break;
    23. 		}
    24. 		if(j == n) c[k ++ ] = b[i];
    25. 	}
    26.     cout << c[0];
    27. 	for(i = 1; i < k; i ++ ){
    28. 		for(j = 0; j < i; j ++ ){
    29. 			if(c[i] == c[j]) break;
    30. 		}
    31. 		if(j == i) cout << " " << c[i];
    32. 	}
    33. 	return 0;
    34. }

    7-85 螺旋方阵

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 15;
    6. int a[N][N];
    7. int main(){
    8.     int n;
    9.     cin >> n;
    10.     int row = 0, col = n - 1, num = 1;
    11.     while(row <= col){
    12.     	for(int i = row; i <= col; i ++ ){
    13.     		a[row][i] = num ++ ;
    14. 		}
    15. 		for(int i = row + 1; i <= col; i ++ ){
    16. 			a[i][col] = num ++ ;
    17. 		} 
    18. 		for(int i = col - 1; i >= row; i -- ){
    19. 			a[col][i] = num ++ ;
    20. 		}
    21. 		for(int i = col - 1; i >= row + 1; i -- ){
    22. 			a[i][row] = num ++ ;
    23. 		}
    24. 		row ++ , col -- ;
    25. 	}
    26. 	for(int i = 0; i < n; i ++ ){
    27. 		for(int j = 0; j < n; j ++ ){
    28. 			printf("%3d", a[i][j]);
    29. 		}
    30. 		puts("");
    31. 	}
    32. 	return 0;
    33. }

    7-88 统计字符出现次数 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 350;
    6. int a[N];
    7. int main(){
    8.     char c;
    9.     c = getchar();
    10.     while(c != '\n'){
    11.         a[c] ++ ;
    12.         c = getchar();
    13.     }
    14.     c = getchar();
    15.     cout << a[c];
    16. 	return 0;
    17. }

    7-92 字符串转换成十进制整数

     感觉就是有没有做过和看看ASCII的问题

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. typedef long long LL;
    5. const int N = 1e5 + 10;
    6. LL fun(char *str){
    7.     LL x = 0;
    8. 	for(int i = 0; i < strlen(str); i ++ ){
    9. 		if(str[i] >= '0' && str[i] <= '9')
    10. 		    x = x * 16 + str[i] - '0';
    11.         else if(str[i] >= 'A' && str[i] <= 'F')
    12. 		    x = x * 16 + str[i] - 'A' + 10;
    13. 	}
    14.     return x;
    15. }
    16. int main(){
    17. 	char s[N], str[N];;
    18. 	cin >> s;
    19. 	int flag = 0, mul = 0, cnt = 0;
    20. 	for(int i = 0; i < strlen(s); i ++ ){
    21. 		s[i] = toupper(s[i]);
    22. 		if(flag == 0){
    23. 			if(s[i] == '-'){
    24.                 flag = 1;
    25.                 mul = 1;
    26.             }
    27. 			else if((s[i] >= '0' && s[i] <= '9') || (s[i] >= 'A' && s[i] <= 'F')){
    28.                 flag = 1;
    29.                 str[cnt ++ ] = s[i];
    30.             }
    31. 		}
    32. 		else{
    33. 			if((s[i] >= '0' && s[i] <= '9') || (s[i] >= 'A' && s[i] <= 'F')){
    34.                 str[cnt ++] = s[i];
    35. 			}
    36. 		}
    37. 	}
    38. 	LL ans = fun(str);
    39. 	if(mul == 1) ans *= -1;
    40.     cout << ans << endl;
    41. 	return 0;
    42. }

    7-103 查找书籍

    一开始忘记map还有自动根据key值排序的功能了 

    1. #include 
    2. #define INF 0x3f3f3f3f
    3. #define x first
    4. #define y second
    5. using namespace std;
    6. typedef long long LL;
    7. const int N = 1e5 + 10;
    8. map<double, string> mp;
    9. int main(){
    10. 	int n;
    11. 	cin >> n;
    12.     getchar();
    13. 	while(n -- ){
    14. 	 	string name;
    15. 	 	double price;
    16. 	 	getline(cin, name);
    17. 	 	cin >> price;
    18.         getchar();
    19. 	 	mp[price] = name;
    20. 	 }
    21. 	 cout << fixed << setprecision(2) << (--mp.end())->x << ", " << (--mp.end())->y << endl;
    22. 	 cout << fixed << setprecision(2) << mp.begin()->x << ", " << mp.begin()->y << endl;
    23. 	return 0;
    24. }

    7-105 平面向量加法 

    一开始nt了一下，进位考虑应该考虑保留最后一位的后一位 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 1e5 + 10;
    8.  
    9. int main(){
    10. 	double x1, x2, y1, y2;
    11. 	cin >> x1 >> y1 >> x2 >> y2;
    12. 	double x = x1 + x2;
    13. 	double y = y1 + y2;
    14.     if(fabs(x) < 0.05) x = 0.0;
    15.     if(fabs(y) < 0.05) y = 0.0;
    16. 	printf("(%.1lf, %.1lf)", x, y);
    17. 	return 0;
    18. }

     7-108 英文单词排序

    哥们觉得vector pair auto 用法需要记一下 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 1e5 + 10;
    8. typedef pair<int, int> pr;
    9. vector v;
    10. int cnt;
    11. string t[N];
    12. int main(){
    13. 	string s;
    14. 	while(cin >> s && s != "#"){
    15. 		v.push_back({s.size(), cnt});
    16. 		t[cnt ++ ]  = s;
    17. 	}
    18. 	sort(v.begin(), v.end());
    19. 	for(auto i : v){
    20. 		cout << t[i.y] << " ";
    21. 	}
    22. 	return 0;
    23. }

    7-109 藏头诗

    这道题上次做还是int行，改成UTF-8把我干蒙了，触及到只是盲区了

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 1e5 + 10;
    8.  
    9. int main(){
    10. 	string s;
    11. 	for(int i = 0; i < 4; i ++ ){
    12. 		getline(cin, s);
    13. 		cout << s[0] << s[1] << s[2];
    14. 	}
    15. 	return 0;
    16. }

    7-110 自动售货机 

     一开始没注意，被以前的思维带偏了，我为什么要想他一样的物品剩多少匹配多少的问题，NND

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 20;
    8. int a[N];
    9. int sum, total;
    10. int main(){
    11. 	int n;
    12. 	while(cin >> n, n != -1){
    13. 		total += n;
    14. 	}
    15. 	while(cin >> n, n != -1){
    16. 		a[n] ++ ; //编号为n的总数量 
    17. 		if(n > 0 && n < 4) sum ++ ;
    18. 		else if(n > 3 && n < 6) sum += 2;
    19. 		else if(n > 5 && n < 9) sum += 3;
    20. 		else sum += 4;
    21. 	}
    22. 	if(total < sum) cout << "Insufficient money" << endl;
    23. 	else{
    24. 		printf("Total:%dyuan,change:%dyuan\n", total, total - sum);
    25. 		for(int i = 0; i < 11; i ++ ){
    26. 			if(a[i]){
    27. 				switch(i){
    28. 					case 1:printf("Table-water:%d;", a[i]);break;
    29. 					case 2:printf("Table-water:%d;", a[i]);break;
    30. 					case 3:printf("Table-water:%d;", a[i]);break;
    31. 					case 4:printf("Coca-Cola:%d;", a[i]);break;
    32. 					case 5:printf("Milk:%d;", a[i]);break;
    33. 					case 6:printf("Beer:%d;", a[i]);break;
    34. 					case 7:printf("Orange-Juice:%d;", a[i]);break;
    35. 					case 8:printf("Sprite:%d;", a[i]);break;
    36. 					case 9:printf("Oolong-Tea:%d;", a[i]);break;
    37. 					case 10:printf("Green-Tea:%d;", a[i]);break;
    38. 				}
    39. 			}
    40. 			
    41. 		} 
    42. 	}
    43. 	return 0;
    44. }

    7-111 停车场管理 

    怪，这题真的怪，代码好像可以卡bug 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 11;
    8. int place[N], tim[N], wait[N];
    9. int waiti = -1;
    10. int main(){
    11. 	int n;
    12. 	cin >> n;
    13. 	while(1){
    14. 		char x;
    15. 		int carnum, cartime;
    16. 		cin >> x >> carnum >> cartime;
    17. 		if(x == 'E') break; 
    18. 		else if(x == 'A'){
    19. 			int flag = 0;
    20. 			for(int i = 1; i <= n; i ++ ){
    21. 				if(place[i] == 0){
    22. 					flag = i;
    23. 					break;
    24. 				}
    25. 			}
    26. 			if(flag){
    27. 				place[flag] = carnum;
    28. 				cout << "car#" << carnum << " in parking space #" << flag << endl;
    29. 			}
    30. 			else{
    31. 				wait[++ waiti] = carnum;
    32. 				cout << "car#" << carnum << " waiting" << endl;
    33. 			}
    34. 		}
    35. 		else if(x == 'D'){
    36. 			int flag = 0;
    37. 			for(int i = 0; i <= n; i ++ ){
    38. 			    if(place[i] == carnum){
    39. 			    	flag = i;
    40. 			        break;
    41. 				} 
    42. 			}
    43. 			if(!flag) cout << "the car not in park" << endl;
    44. 			else{
    45. 				cout << "car#" << carnum << " out,parking time " << tim[flag] << endl;
    46. 				place[flag] = 0;
    47. 				tim[flag] = 0;
    48. 				int t = 0;
    49. 				for(int i = flag; i <= n; i ++ ){
    50. 					place[i] = place[i + 1];
    51. 					tim[i] = tim[i + 1];
    52. 					if(i == n){
    53. 						place[i] = 0;
    54. 						tim[i] = 0;
    55. 					}
    56. 					t = i;
    57. 				}
    58. 				if(waiti != -1){
    59. 					place[t] = wait[0];
    60. 					for(int i = 0; i < waiti; i ++ ){
    61. 						wait[i] = wait[i + 1];
    62. 					}
    63. 					waiti -- ;
    64. 					cout << "car#" << place[t] << " in parking space #" << t << endl;
    65. 				}
    66. 				
    67. 			}
    68. 		}
    69. 		for(int i = 1; i <= n; i ++ ){
    70. 			if(place[i]) tim[i] ++ ;
    71. 		}
    72. 	}
    73. 	return 0;
    74. }

    7-112 值班安排 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 20;
    8. struct st{
    9.     int obj1;
    10.     int obj2;
    11.     int num;
    12.     int ret;
    13. }st[N];
    14. int judge(int *d, int n){
    15.     int ret = 1;
    16.     for(int i = 0; i < n; i ++ ){
    17.         if(st[i].ret){
    18.             if(d[st[i].obj1] != st[i].num - 1){
    19.                 ret = 0;
    20.                 break;
    21.             }
    22.         }
    23.         else if(d[st[i].obj1] + st[i].num != d[st[i].obj2]){
    24.             ret = 0;
    25.             break;
    26.         }
    27.     }
    28.     return ret;
    29. }
    30. int main(){
    31. 	int n;
    32. 	cin >> n;
    33.     getchar();
    34.     string s;
    35. 	for(int i = 0; i < n; i ++ ){
    36.         getline(cin, s);
    37.         st[i].obj1 = s[0] - 'A';
    38.         if(s[1] == '='){
    39.             st[i].ret = 1;
    40.             st[i].num = s[2] - '0';
    41.         }
    42.         else{
    43.             st[i].ret = 0;
    44.             st[i].obj2 = s[2] - 'A';
    45.             st[i].num = s[3] - '0';
    46.             if(s[1] == '<') st[i].num *= -1;
    47.         }
    48.     }
    49.     int day[7];
    50.     int a, b, c, d, e, f, g;
    51.     for(a = 0; a < 7; a ++ ){
    52.         for(b = 0; b < 7; b ++ ){
    53.             for(c = 0; c < 7; c ++ ){
    54.                 for(d = 0; d < 7; d ++ ){
    55.                     for(e = 0; e < 7; e ++ ){
    56.                         for(f = 0; f < 7; f ++ ){
    57.                             for(g = 0; g < 7; g ++ ){
    58.                                 day[0] = a;
    59.                                 day[1] = b;
    60.                                 day[2] = c;
    61.                                 day[3] = d;
    62.                                 day[4] = e;
    63.                                 day[5] = f;
    64.                                 day[6] = g;
    65.                                 if(judge(day, n)) goto E;
    66.                             }
    67.                         }
    68.                     }
    69.                 }
    70.             }
    71.         }
    72.     }
    73. E:
    74.     if(judge(day, n)){
    75.         char Day[7];
    76.         for(int i = 0; i < 7; i ++ ){
    77.             Day[day[i]] = i + 'A';
    78.         }
    79.         Day[7] = '\0';
    80.         puts(Day);
    81.     }
    82. 	return 0;
    83. }

    7-113 完美的代价

    Impossible有两个条件：

    1.当n为奇数，如果有两个及以上不能配对的，说明不符合回文串条件

    2.当n为偶数，如果相同数字出现奇数次 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 11;
    8. int main(){
    9. 	int n;
    10. 	string s;
    11. 	cin >> n >> s;
    12. 	int ed = n - 1;
    13. 	int step = 0;
    14. 	bool flag = false;
    15. 	for(int i = 0; i < ed; i ++ ){
    16. 		for(int j = ed; j >= i; j -- ){
    17. 			if(i == j){
    18. 				if(n % 2 == 0 || flag == true){
    19. 					cout << "Impossible"  << endl;
    20. 					return 0;
    21. 				}
    22. 				flag = true;
    23. 				step += n / 2 - i;//当前位置到中间位置需要的步数
    24. 			}
    25. 			else if(s[i] == s[j]){
    26. 				for(int k = j; k < ed; k ++ ){
    27. 					swap(s[k], s[k + 1]);
    28. 					step ++ ;
    29. 				}
    30. 				ed -- ;
    31.                 break;//重点
    32. 			}
    33. 		}
    34. 	}
    35. 	cout << step << endl;
    36. 	return 0;
    37. }

    L2-1 盲盒包装流水线 

    一开始只有22分，卡在了开的不够大，紫砂 ，现在st开的数据我不知道具体怎么算的 QAQ，感觉·就看样例的话就是N=可能的最大数据 / S本题样例5 就不会溢出了（

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 1e5 + 10;
    8. stack<int> st[50010];
    9. queue<int> q;
    10. int a[N];
    11. int main(){
    12. 	int n, s, k;
    13. 	cin >> n >> s;
    14. 	for(int i = 1; i <= n; i ++ ){
    15. 		int x;
    16. 		cin >> x;
    17. 		q.push(x);
    18. 	}
    19. 	
    20. 	int col = n / s;
    21. 	for(int i = 0; i < col; i ++ ){
    22. 		for(int j = 0; j < s; j ++ ){
    23. 			int x;
    24. 			cin >> x;
    25. 			st[i].push(x);
    26. 		}
    27. 		for(int j = 0; j < s; j ++ ){
    28. 			int x = q.front();
    29. 			q.pop();
    30. 			int y = st[i].top();
    31. 			st[i].pop();
    32. 			a[x] = y;
    33. 		}
    34. 	}
    35. 	cin >> k;
    36. 	while(k -- ){
    37. 		int x;
    38. 		cin >> x;
    39. 		if(a[x] == 0) cout << "Wrong Number" << endl;
    40. 		else cout << a[x] << endl;
    41. 	}
    42. 	return 0;
    43. }

    L2-2 点赞狂魔 

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 110;
    8. struct person{
    9. 	string name;
    10. 	set tag;
    11. 	double ave;
    12. }p[N];
    13. bool cmp(person x, person y){
    14. 	if(x.tag.size() == y.tag.size()) return x.ave < y.ave;
    15. 	return x.tag.size() > y.tag.size();
    16. }
    17. int main(){
    18. 	int n;
    19. 	cin >> n;
    20. 	for(int i = 0; i < n; i ++ ){
    21. 		string s;
    22. 		cin >> s;
    23. 		p[i].name = s;
    24. 		int k;
    25. 		cin >> k;
    26. 		for(int j = 0; j < k; j ++ ){//woc用while循环过不了
    27. 			LL x;
    28. 			cin >> x;
    29. 			p[i].tag.insert(x);
    30. 		}
    31. 		p[i].ave = 1.0 * k / p[i].tag.size();
    32. 	}
    33. 	sort(p, p + n, cmp);
    34. 	if(n == 0) cout << "- - -" << endl;
    35. 	else if(n == 1) cout << p[0].name << " - -" << endl;
    36. 	else if(n == 2) cout << p[0].name << " " << p[1].name << " -" << endl;
    37. 	else cout << p[0].name << " " << p[1].name << " " << p[2].name << endl;
    38. 	
    39. 	return 0;
    40. }

    L2-3 浪漫侧影

    1. #include 
    2. using namespace std;
    3. #define INF 0x3f3f3f3f
    4. #define x first
    5. #define y second
    6. typedef long long LL;
    7. const int N = 1e5 + 10;
    8. int midt[25], postt[25], tree[N], level;
    9. void createTree(int *postt, int *midt, int n, int idx){
    10. 	if(!n) return ;
    11. 	int k = 0;
    12. 	while(postt[n - 1] != midt[k]) k ++ ;
    13. 	tree[idx] = midt[k];
    14. 	level = max(level, idx);
    15. 	createTree(postt, midt, k, idx * 2);
    16. 	createTree(postt + k, midt + k + 1, n - k - 1, idx * 2 + 1);
    17. }
    18. int main(){
    19. 	int n;
    20. 	cin >> n;
    21. 	for(int i = 0; i < n; i ++ ) cin >> midt[i];
    22. 	for(int i = 0; i < n; i ++ ) cin >> postt[i];
    23. 	createTree(postt, midt, n, 1);
    24. 	for(int i = 1; i <= 22; i ++ ){
    25. 		if(level < pow(2, i)){
    26. 			level = i;
    27. 			break;
    28. 		}
    29. 	}
    30. 	cout << "R: ";
    31. 	int cnt = 0, idx = 1;
    32. 	while(cnt < level){
    33. 		if(tree[idx]) cout << tree[idx];
    34. 		if(tree[idx * 2 + 1]) idx = idx * 2 + 1;
    35. 		else if(tree[idx * 2]) idx *= 2;
    36. 		else{
    37. 			idx *= 2;
    38. 			while(!tree[idx]) idx -- ;
    39. 		}
    40. 		cnt ++ ;
    41. 		if(cnt != level) cout << ' ';
    42. 		else cout << endl;
    43. 	}
    44. 	cout << "L: ";
    45. 	cnt = 0, idx = 1;
    46. 	while(cnt < level){
    47. 		if(tree[idx]) cout << tree[idx];
    48. 		if(tree[idx * 2]) idx *= 2;
    49. 		else if(tree[idx * 2 + 1]) idx = idx * 2 + 1;
    50. 		else{
    51. 			idx *= 2;
    52. 			while(!tree[idx]) idx ++ ;
    53. 		}
    54. 		cnt ++ ;
    55. 		if(cnt != level) cout << ' ';
    56. 		else cout << endl;
    57. 	}
    58. 	return 0;
    59. }

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  • 原文地址：https://blog.csdn.net/weixin_63914060/article/details/126932992