1060 Are They Equal

1060 Are They Equal

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123×105 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line YES if the two numbers are treated equal, and then the number in the standard form 0.d[1]...d[N]*10^k (d[1]>0 unless the number is 0); or NO if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

总结：没有全部写出来，只有19分

代码：

①：最后一个测试点是如果读取的位数

②：如果a=123.45 b=12 这个代码也是会打印出 YES（判断是否相等出错）

#include 
using namespace std;
 
int n;
void count(string a){
    if(a=="0"){
        printf(" 0.0*10^0");
        return;
    }
    bool st=false;//表示是否遇到了数字
    string s;
    int cot=0,num=0;//cot表示指数，num表示数字的有效位数
    if(a[0]=='0'){
        for(int i=2;isize() && num
            if(a[i]=='0' && st==false)   cot++;
            else{
                st=true;
                s+=a[i];
                num++;
            }
        }
    }
    else{
        for(int i=0;isize();i++){
            if(a[i]=='.'){
                st=true;
                continue;
            }
            if(!st) cot++;
            if(num
                num++;
                s+=a[i];
            }
        }
    }
    printf(" 0.");
    cout << s;
    printf("*10^%d",cot);
    return;
}
 
int main(){
    string a,b;
    cin >> n >> a >> b;
    
    bool flag=false;
    for(int i=0;isize() && isize() && i
        if(a[i]!=b[i])  flag=true;
    
    if(!flag){
        printf("YES");
        count(a);
    }
    else{
        printf("NO");
        count(a);count(b);
    }
    
    return 0;
}

 大佬代码：
 

#include 
#include 
using namespace std;
int main() {
    int n, p = 0, q = 0;
    char a[10000], b[10000], A[10000], B[10000];
    scanf("%d%s%s", &n, a, b);
    int cnta = strlen(a), cntb = strlen(b);
    //记录小数点的位置 
    for(int i = 0; i < strlen(a); i++) {
        if(a[i] == '.') {
            cnta = i;
            break;
        }
    }
    for(int i = 0; i < strlen(b); i++) {
        if(b[i] == '.') {
            cntb = i;
            break;
        }
    }
    int indexa = 0, indexb = 0;
    //找到第一个有效数字（即非0且不是小数点的位置） 
    while(a[p] == '0' || a[p] == '.') p++;
    while(b[q] == '0' || b[q] == '.') q++;
    //找到次方的大小 
	if(cnta >= p)
        cnta = cnta - p;
    else
        cnta = cnta - p + 1;
    if(cntb >= q)
        cntb = cntb - q;
    else
        cntb = cntb - q + 1;
    if(p == strlen(a))
        cnta = 0;
    if(q == strlen(b))
        cntb = 0;
    //从有效数字位开始读取数字 
    while(indexa < n) {
    	//如果没有超出数的范围且不是小数点 
        if(a[p] != '.' && p < strlen(a))
            A[indexa++] = a[p];
        //如果超过了数的范围则补0 
        else if(p >= strlen(a))
            A[indexa++] = '0';
        p++;
    }
    while(indexb < n) { 
        if(b[q] != '.' && q < strlen(b))
            B[indexb++] = b[q];
        else if(q >= strlen(b))
            B[indexb++] = '0';
        q++;
    }
    if(strcmp(A, B) == 0 && cnta == cntb)
        printf("YES 0.%s*10^%d", A, cnta);
    else
        printf("NO 0.%s*10^%d 0.%s*10^%d" , A, cnta, B, cntb);
    return 0;
}

好好学习，天天向上！

我要考研！