PAT-1060 Are They Equal （科学计数法）

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...dN*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

题目大意：给出两个数，要求判断两数在保留小数点后n位并以科学计数法表示时是否相等。

主要思路：该题的核心步骤主要是找出一个数的小数点位置（没有则设置为字符串长度），第一个有效数位置（非0和非小数点），然后根据这两个值计算出指数值。在比较的时候应分两部分，底数和指数，都相等时才相等，其中底数的小数点后n位部分存于另一目标数组。（注意：诸如0.000的为特殊形式，有效位置等于字符串长度，此时应把指数设为0，防止出现0.00与0.000不等的错误）

#include <cstdio>
#include <string.h>
void print_str(char * s, int n, int count);
int main(void) {
    int n, i;
    char s1[105], s2[105], res1[105], res2[105];
    
    scanf("%d %s %s", &n, s1, s2);
    int count1 = strlen(s1), count2 = strlen(s2);       //小数点位置
    int p = 0, q = 0;                                   //第一个有效数字的位置
    
    //分别计算两个数的 小数点位置，首个有效数位置，幂指数
    for (i = 0; i < strlen(s1); i++) {
        if (s1[i] == '.') {
            count1 = i;
            break;
        }           
    }
    for (i = 0; i < strlen(s2); i++) {
        if (s2[i] == '.') {
            count2 = i;
            break;
        }            
    }
    while (s1[p] == '0' || s1[p] == '.')    p++;
    while (s2[q] == '0' || s2[q] == '.')    q++;
    int k1 = count1 - p;                                //k1, k2 表示指数
    if (k1 < 0)             k1++;
    int k2 = count2 - q;
    if (k2 < 0)             k2++;
    if (p == strlen(s1))    k1 = 0;                     //0.000.. 的情况
    if (q == strlen(s2))    k2 = 0;
    
    //将两字符串的n位 底数部分 复制到结果数组
    int index1 = 0, index2 = 0;
    while (index1 < n) {
        if (p < strlen(s1) && s1[p] != '.')
            res1[index1++] =  s1[p];
        else if (p >= strlen(s1))
            res1[index1++] = '0';
        p++;
    }
    res1[index1] = '\0';
    while (index2 < n) {
        if (q < strlen(s2) && s2[q] != '.')
            res2[index2++] =  s2[q];
        else if (q >= strlen(s2))
            res2[index2++] = '0';
        q++;
    }
    res2[index2] = '\0';
    if (!strcmp(res1, res2) && k1 == k2)
        printf("YES 0.%s*10^%d\n", res1, k1);
    else
        printf("NO 0.%s*10^%d 0.%s*10^%d\n", res1, k1, res2, k2);
        
    return 0;
}