A1060. 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.d₁...d_N*10^k" (d₁>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

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 using namespace std;  // 0000.0012   0123.34
 char str1[], str2[];
 int process(char s[], int len){
     int i,j, exp;
     for(i = ; s[i] != '\0' && s[i] == ''; i++);  //去除前导0
     if(s[i] != '\0'){
         for(j = ; s[j + i] != '\0'; j++)
             s[j] = s[j + i];
         s[j] = '\0';
     }else{
         for(int k = ; k < len; k++)
             s[k] = '';
         s[len] = '\0';
         return ;
     }
     if(s[] == '.'){  //.0012
         for(i = ; s[i] != '\0' && s[i] == ''; i++);
         if(s[i] == '\0'){
             exp = ;
             s[] = '';
         }else{
             exp = i - ;
             exp *= -;
             for(j = ; s[i + j] != '\0'; j++)
                 s[j] = s[j + i];
             s[j] = '\0';
         }
     }else{  //123.456
         for(i = ; s[i] != '\0' && s[i] != '.'; i++);
         if(s[i] == '\0'){
             exp = i;
         }else{
             exp = i;
             for(; s[i + ] != '\0'; i++)
                 s[i] = s[i + ];
             s[i] = '\0';
         }
     }
     for(j = ; s[j] != '\0'; j++);
     while(j < len){
         s[j++] = '';
     }
     s[len] = '\0';
     return exp;
 }
 int main(){
     int N, exp1, exp2;
     scanf("%d %s %s", &N, str1, str2);
     exp1 = process(str1, N);
     exp2 = process(str2, N);
     int tag = ;
     for(int i = ; str1[i] != '\0'; i++)
         if(str1[i] != str2[i]){
             tag = ;
             break;
         }
     if(tag ==  && exp1 == exp2){
         printf("YES 0.%s*10^%d", str1, exp1);
     }else{
         printf("NO 0.%s*10^%d 0.%s*10^%d", str1, exp1, str2, exp2);
     }
     cin >> N;
     return ;
 }

总结：

1、题意：将给出的两个数字变成保留指定位小数的科学计数法的数字，之后看其是否相等。

2、主要分为两种数字，大于1和小于1（0.000123， 1234.5678），其次要注意有0000123的情况出现，要先去除前导0.

3、主要要做的就是想办法求出不带小数点的且符合要求的底数。

4、测试样例： 3 0.0 0

输出：YES 0.000*10^0