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1、n项多项式求和
s=1/1x2/1…1/(2n-1)x(2n/(2n-1))
#include#include #include #include #define N 5 int main() { int n; float term=1,sum=0.0;//初始化term和s的值 for(n=1;term>1e-6;n++){ //精度控制在le-6 term=(1.0/(2*n-1))*(2*n/(2*n-1)); sum+=term; } printf("%f",sum); return 0; } - 1
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2、nn魔方矩阵实现每行、每列、每一对角线上的元素之和相等 (将1~nn每一个元素放到合适位置)
#include#include #include #include #define N 15 int main() { int a[N][N]= {0}; int k; printf("请输入魔方矩阵k="); scanf("%d",&k); while(1) { if(k%2==0) { printf("魔方阶数只能为奇数,请重新输入\n"); scanf("%d",&k); } else { break; } } int i,j,m,n;//m、n用于暂时存放上一次i、j i=0,j=k/2; a[i][j]=1;//初始位置为第一行、中间列 for(int x=2; x<=k*k; x++) { m=i,n=j; i--,j++;//向45°右上方移动也就是行减1,列加1 if(i<0)i=k-1;//如果已经在第一行,那么下一次移动到最后一行 if(j>=k)j=0; //如果已经在第n列,下一次移动到第一列 if(a[i][j]!=0) { //如果该位置已经有元素占领,则存放在下个位置 i=m+1; j=n; } a[i][j]=x; } for(int i=0; i<k; i++) { for(int j=0; j<k; j++) { printf("%d ",a[i][j]); } printf("\n"); } return 0; } - 1
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3、 求前20项和 2/1+3/2+5/3…13/8+21/13
#include#include #include #include #define N 15 int main() { int i,n=20; float a=2,b=1,s=0,t;//a是分子,b是分母 for(int i=0;i<n;i++){ s+=(a/b); t=a; a+=b;//分子为前一项分子和分母之和 b=t; //分母为前一项分子 } printf("%f",s);//32.660263 return 0; } - 1
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4、第一项为2 第二项为3
序列最后两项乘积<10,则最后项为两项乘积
如果最后两项乘积>10,则拆分十位和个位,分别位新的两项#include#include #include #include #define N 15 int main() { int n,cnt,i=2; int a[100]; a[0]=2,a[1]=3; while(cnt!=100) { int t=a[i-1]*a[i-2]; if(t<10) { a[i++]=t; cnt+=1; } else { a[i++]=t/10; a[i++]=t%10; cnt+=2; } } for(int i=0;i<100;i++){ printf("%d ",a[i]); } return 0; } - 1
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5、调和级数 1/1+1/2+1/3+…+1/n 的前n项和,要求用分数
#include#include #include #include #define N 15 int gcb(int x,int y) { return y==0?x:gcb(y,x%y); } int main() { int n,u=1,v=1;//n为级数项数、u为分子、v为分母 scanf("%d",&n); for(int i=2; i<=n; i++) { //通分求和 u=u*i+1*v; v*=i; int g=gcb(u,v);//用最大公约数,约分 u/=g; v/=g; printf("%d/%d ",u,v); } return 0; } - 1
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6、求出sinx的近似值 sinx用级数展开
#include#include #include #include #define N 15 int func(int n) { int k=1; for(int i=1; i<=n; i++) { k=k*i; } return k; } int main() { int x,n,sign=1; double term,s=0; scanf("%d",&x); term=x; for(n=1; term>5e-6; n+=2) { //每一项为奇数项 term=pow(x,n)/func(n); s=s+sign*term; sign=-sign; } printf("sin%d的近似值为: %.8lf",x,s); return 0; } - 1
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求级数 1/1-1/2+1/3+…+1/n
#include#include #include #include #define N 15 int main() { int n,sign=1;//设置负号变量 scanf("%d",&n); float s=0; for(int i=1;i<=n;i++){ s+=(float)sign/i; sign=-sign; } printf("%f",s); return 0; } - 1
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7、输出分母小于或者等于n的最简真分数(不能约分),并且从小到大排序输出
n=5 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5
#include#include #include #include #define maxn 999 struct struct_result { int numerator;//分子 int denominator;//分母 float result;//分数 bool operator <(const struct_result x) { return result<=x.result; } } list[maxn]; int length; int gcb(int x,int y) { return y==0?x:gcb(y,x%y); } //快排 int partiton(struct_result list[],int low,int high) { struct_result pivot=list[low]; while(low<high){ while(low<high&&list[high].result>=pivot.result)high--; list[low]=list[high]; while(low<high&&list[low].result<=pivot.result)low++; list[high]=list[low]; } list[low]=pivot; return low; } void quick_sort(struct_result list[],int low,int high) { if(low<high) { int pivotPos=partiton(list,low,high); quick_sort(list,low,pivotPos-1); quick_sort(list,pivotPos+1,high); } } int main() { int n; scanf("%d",&n); for(int i=1; i<=n; i++) { //枚举分子和分母 for(int j=i+1; j<=n; j++) { if(gcb(i,j)==1) { //如果为最简真分数则添加进入数组 list[length].numerator=i; list[length].denominator=j; list[length++].result=(float)i/j; } } } quick_sort(list,0,length-1); for(int i=0; i<length; i++) { printf("%d/%d ",list[i].numerator,list[i].denominator); if(i+1%10==0)printf("\n");//每十个换行 } return 0; } - 1
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8、对100个点相互组合组成的线段求出距离最小的长度
#include#include #include #include #define maxn 999 struct point{ int x,y; }point[maxn]; int dis(struct point a,struct point b){ return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); } int main() { for(int i=0;i<100;i++){ scanf("%d%d",&point[i].x,&point[i].y); } int mind=0x7f7f7f7f; for(int i=0;i<100;i++) for(int j=i+1;j<100;j++){ if(dis(point[i],point[j])<mind){ mind=dis(point[i],point[j]); } } printf("maxdis=",sqrt(mind)); return 0; } - 1
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9、给出100个点,任意三个点组成三角形,求这些三角形的最大面积是多少
给出三个点,可以用向量来做,利用叉积=|a|x|b|xsina 等于平行四边面积为对于三角型面积二倍来求出三角形
也可以利用海伦公式,求出周长为2p 利用海伦公式s=根号下px(p-x)x(p-y)x(p-z)
#include#include #include #include #define maxn 999 typedef struct Point{ double x,y; }Point; Point decPoint(Point p1,Point p2){ //将坐标转换为向量 Point ret; ret.x=p1.x-p2.x; ret.y=p1.y-p2.y; return ret; } double multiPoint(Point p1,Point p2){//叉乘=x2y1-x1y2 return p1.x*p2.y-p1.y*p2.x; } double areaByThreePoint(Point A,Point B,Point C){ //三角形面积 return fabs(multiPoint(decPoint(A,B),decPoint(B,C))); } int main() { int i,j,k; double maxarea; Point pts[110]; for(int i=0;i<100;i++){ scanf("%lf%lf",&pts[i].x,&pts[i].y); } maxarea=areaByThreePoint(pts[0],pts[1],pts[2]); for(i=0;i<100;i++){ for(j=i+1;j<100;j++){ for(k=j+1;k<100;k++){ double t=areaByThreePoint(pts[i],pts[j],pts[k]); if(maxarea<t)maxarea=t; } } } printf("面积最大为:%.2lf",maxarea/2.0); return 0; } - 1
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10、给出100个圆的圆心坐标和半径,判断哪些圆相切,将相切的圆用一个链表相连接,链表中存放圆的编号
#include#include #include #include #define maxn 999 typedef struct node{ int m,n; struct node* next; }Node; typedef struct circle{ int x,y; double r; }Circle; bool isQie(circle a,circle b){ double dis=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);//圆心直接距离的平方 if(pow(a.r+b.r,2)==dis)return true; return false; } int main() { Node* head,p; Circle cir[110]; for(int i=0;i<100;i++){ scanf("%d%d%f",cir[i].x,cir[i].y,cir[i].r); } for(int i=0;i<100;i++){ for(int j=i+1;j<100;j++){ if(isQie(cir[i],cir[j])){ p=(Node*)malloc(sizeof(Node));//头插法 p->m=i; p->n=j; p->next=head; head=p; } } } return 0; } - 1
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11、求解二元一次方程 ax^2+bx+c=0
#include#include #include #include #define maxn 999 int main() { double a,b,c,disc,x1,x2; scanf("%lf%lf%lf",&a,&b,&c); if(fabs(a)<1e-6)printf("不是二元一次方程");//a=0 else{ disc=b*b-4*a*c;//判断几个实根 if(fabs(disc)<=1e-6)printf("%f",-b/(2*a));//只有一个根 else if(fabs(disc)>1e-6){ x1=(-b+sqrt(disc))/(2*a); x2=(-b-sqrt(disc))/(2*a); printf("两个解分别为%lf,%lf",x1,x2); }else{ printf("无解");//disc<0 } } return 0; } - 1
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12、利用二分法找到 2x^3- 4x^3+3x-6=0 位于-10~10直接的一个跟 x=3
#include#include #include #include #define eps 1e-6 double f(double x){ return 2*x*x*x-4*x*x+3*x-6; } int main() { double m=-10,n=10; double i,s,k; if(f(m)*f(n)<0){ while(fabs(m-n)>eps){//两个端点不相同 i=(m+n)/2; s=f(i); if(fabs(s)<eps)break; else if(f(i)*f(m)<0)n=i; else if(f(i)*f(n)<0)m=i; } } printf("%lf%lf",m,n); k=(m+n)/2; printf("%lf",k); return 0; } - 1
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13、利用牛顿迭代法求12题根,x1=1.5
#include#include #include #include #define eps 1e-6 float fun(float x){ return 2*x*x*x-4*x*x+3*x-6; } float fun1(float x){ return 6*x*x-8*x+3; } int main() { float x1=1.5,x2=x1-fun(x1)/fun1(x1); //根据斜率公式表示新的与x轴交点x2 while(fabs(x1-x2)>1e-8){ x1=x2; x2=x1-fun(x1)/fun1(x1); } printf("%f",x2); return 0; } - 1
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14、判断某点是否在三角形内部
利用叉乘,判断D点是否在AB BC CA 的左侧, 分别CA 与AD做叉乘,AB 与BD做叉乘,BC与CD做叉乘,判断符号是否相同
#include#include #include #include #define eps 1e-6 typedef struct Point{ double x,y; }Point; Point decPoint(Point p1,Point p2){ //定义向量 Point ret; ret.x=p1.x-p2.x; ret.y=p1.y-p2.y; return ret; } double multiPoint(Point ver1,Point ver2){ //叉乘 return (ver1.x*ver2.y-ver1.y*ver2.x); } int sign(double x){ //判断符号 if(fabs(x)<eps)return 0;//等于0 else if(x>0)return 1; return 0; } int main() { Point points[4];// A B C D for(int i=0;i<4;i++){ scanf("%lf%lf",&points[i].x,&points[i].y); } int sgn[3];//存放三个叉乘后向量的符号 //叉乘向量 CA 和 AD Point CA=decPoint(points[0],points[2]); Point AD=decPoint(points[3],points[0]); double verA=multiPoint(CA,AD); sgn[0]=sign(verA); printf("%lf\n",verA); //叉乘 AB BD Point AB=decPoint(points[1],points[0]); Point BD=decPoint(points[3],points[1]); double verB=multiPoint(AB,BD); sgn[1]=sign(verB); printf("%lf\n",verB); //叉乘 BC CD Point BC=decPoint(points[2],points[1]); Point CD=decPoint(points[3],points[2]); double verC=multiPoint(BC,CD); sgn[2]=sign(verC); printf("%lf\n",verC); if(sgn[0]==0||sgn[1]==0||sgn[2]==0)printf("Point D is on the triangle!"); //在边上或者延长线上 else if(sgn[0]==sgn[1]&&sgn[1]==sgn[2]) printf("Point D is in the triangle!"); else printf("Point D is out of the triangle!"); return 0; } - 1
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15、递归编写函数,求出数组中所有数的和
#include#include #include #include #define eps 1e-6 int sum(int n,int a[]){ if(n==1)return a[0]; return sum(n-1,a)+a[n-1]; } int main() { int a[10]={1,2,3,4,5,6,7,8,9,10}; printf("%d",sum(10,a)); return 0; } - 1
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16、利用递归求出一个整数个个位数之和
#include#include #include #include #define eps 1e-6 int sum(int n){ if(n<10)return n; return n%10+sum(n/10); } int main() { int n=1031; printf("%d",sum(n)); return 0; } - 1
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17、一根木条在五分之二处截断, 将剩余木条也在五分至二处截断,知道所有木条长度均小于5,求出一个折成多少木条
#include#include #include #include #define eps 1e-6 int f(double n){ if(n<=5)return 1; return f(2*n/5)+f(3*n/5); } int main() { printf("%d",f(6)); return 0; } - 1
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18、读入文件中学生成绩,将学生成绩用一个链表有高分到低分排列起来,一边读入,一边排序
#include#include //malloc typedef struct Grade{ int num; char name[20]; int disscore; int repscore; int testscore; int totscore; struct Grade* next; }GradeNode,*GradeList; int main() { FILE* fpin=fopen("2018Exp.txt","r");//打开文件 GradeList head; head->next=NULL; GradeNode* gNode; while(!feof(fpin)){ //如果文件不为空 gNode=(GradeNode*)malloc(sizeof(GradeNode)); fscanf(fpin,"%d",&gNode->num); fscanf(fpin,"%s",&gNode->name); fscanf(fpin,"%d",&gNode->disscore); fscanf(fpin,"%d",&gNode->repscore); fscanf(fpin,"%d",&gNode->testscore); gNode->totscore=(int)(0.2*gNode->disscore+0.2*gNode->repscore+0.6*gNode->testscore); GradeNode* p=head->next;//p为当前节点 GradeNode* q=head;//q为p的前一个节点 while(p&&p->totscore>gNode->totscore){//找到gNode所在合适的位置 p=p->next; q=q->next; } q->next=gNode; gNode->next=p; } fclose(fpin);//关闭文件 return 0; } - 1
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19、从键盘接收学生个数、学生数据返回一个学生链表,学号由小到大排名,将链表写入文件
#include#include //malloc typedef struct Student{ int stuNum; int score; char sex; char name[20]; struct Student* next; }Student,*StuList; StuList head=NULL;//设置头节点 StuList CreateList(int n){ Student* stu; while(n){ stu=(Student*)malloc(sizeof(Student)); scanf("%d%d",&stu->stuNum,&stu->score); scanf("%c",&stu->sex); scanf("%s",&stu->name); Student *p=head->next; Student *q=head; while(p&&p->stuNum<stu->stuNum){ p=p->next; q=q->next; } q->next=stu; stu->next=p; n--; } return head; } bool WriteFile(StuList stuList){ FILE* fpin; if((fpin=fopen("output.txt","w"))==NULL){ printf("FILE Open Error!"); return false; } while(stuList){ fprintf(fpin,"%d ",stuList->stuNum); fprintf(fpin,"%d ",stuList->score); fprintf(fpin,"%c ",stuList->sex); fprintf(fpin,"%s\n ",stuList->name); stuList=stuList->next; } fclose(fpin); return true; } int main() { int n; printf("请输入学生个数: "); scanf("%d",&n); head=CreateList(n); WriteFile(head); return 0; } - 1
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20、将小写字母转换为大写字母,并且写入文件中
#include#include //malloc int main() { FILE *fpin; char str[100]; printf("请输入字符串: \n"); gets(str); int i=0; while(str[i]!='\0'){ if(str[i]>='a'&&str[i]<='z'){ str[i]=str[i]-'a'+'A'; } i++; } if((fpin=fopen("test.txt","w"))==NULL){ printf("无法打开文件\n"); } else{ fputs(str,fpin); fclose(fpin); } return 0; } - 1
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21、创建学生链表,将指定年龄的学生删除,将删除后的链表写入文件
#include#include //malloc typedef struct Student{ int num; int age; char name[20]; struct Student * next; }Student,*StuList; StuList head; StuList CreateList(int n){ //创建带头结点的单链表 Student *p,*r=head; while(n){ //尾插法 p=(Student*)malloc(sizeof(Student)); scanf("%d%d%s",&p->num,&p->age,&p->name); r->next=p; r=p; n--; } r->next=NULL; return head; } void save(StuList list,int age){ FILE *file; if((file=fopen("output.txt","w"))==NULL)printf("File Open Error!"); Student *p=list->next; Student *q=list; while(p){ if(p->age==age){ //删除节点 q=p->next; free(p); p=q->next; } } p=head->next; while(p){ fprintf(file,"%5d\n",p->num); fprintf(file,"%s\n",p->name); fprintf(file,"5d\n",p->age); p=p->next; } fclose(file); } int main() { StuList list=CreateList(10); save(list,20); return 0; } - 1
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22、从键盘读入学生链表,进行冒泡排序,在从文件读入一个学生链表进行冒泡排序,最后将两个链表归并排序
#include#include typedef struct Student{ int num; char name[10]; int grade; struct Student* next; }Student,*StuList; void BubbleSort(StuList list){ //传入带头节点的链表 bool flag=true; Student *p,*q,*r; while(flag){ // q p r 相对关系 q=list; p=list->next; r=p->next; flag=false; while(r){ if(p->grade>r->grade){ p->next=r->next;//交换节点 r->next=p; q->next=r; p=r;//调回相对关系 r=p->next; flag=true; } q=p; p=r; r=r->next; } } } StuList CreateList(int n){ StuList head=(Student*)malloc(sizeof(Student)); head->next=NULL; Student *p; while(n){ p=(Student*)malloc(sizeof(Student)); printf("请输入学生学号: "); scanf("%d",&p->num); printf("请输入学生姓名: "); scanf("%s",&p->name); printf("请输入学生成绩: "); scanf("%d",&p->grade); p->next=head->next; head->next=p;//头插法 n--; } return head; } //按照学生成绩打印信息 void PrintList(StuList list){ StuList p=list->next; while(p){ printf("%d-->",p->grade); p=p->next; } printf("\n"); } StuList ReadList(int n){ FILE *fpin; if((fpin=fopen("in.txt","r"))==NULL){ printf("File Open Error!"); exit(0); } StuList head; Student *p; while(n--){ p=(Student*)malloc(sizeof(Student)); fscanf(fpin,"%d%s%d",&(p->num),&(p->name),&(p->grade)); p->next=head; //尾插法 head=p; } fclose(fpin); return head; } //递归的Merge StuList Merge(Student *pHead1,Student *pHead2){ if(pHead1==NULL){ return pHead2; } if(pHead2==NULL){ return pHead1; } Student* head=(Student*)malloc(sizeof(Student)); if(pHead1->grade<pHead2->grade){ head=pHead1; head->next=Merge(pHead1->next,pHead2); }else{ head=pHead2; head->next=Merge(pHead1,pHead2->next); } return head; } int main() { int n;//学生个数 printf("请输入学生个数: "); scanf("%d",&n); StuList list1=CreateList(n); PrintList(list1); BubbleSort(list1); printf("请输入学生个数: "); scanf("%d",&n); StuList list2=CreateList(n); PrintList(list2); BubbleSort(list2);//冒泡排序 printf("Merge两个链表\n"); StuList list=(Student*)malloc(sizeof(Student)); list->next=Merge(list1->next,list2->next); PrintList(list); return 0; } - 1
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- 原文地址：https://blog.csdn.net/weixin_46503238/article/details/126532741
