PAT(甲级)渡劫(二十)-Are They Equal(25)

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PAT(甲级)渡劫(二十)-Are They Equal(25)

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 200;
char s1[maxn],s2[maxn];
int n;

void solve(char *str,char *ans,int n,int &e){
    int len = strlen(str);
    int point = len;    // 小数点的索引
    for(int i = 0 ; i < len ; i++){
        if(str[i] == '.'){
            point = i;
            break;
        }
    }
    int not0 = len; // 第一个非0的索引
    for(int i = 0 ; i < len ; i++){
        if(str[i] != '.' && str[i] != '0'){
            not0 = i;
            break;
        }
    }
    // 数字为0
    if(not0 == len){
        e = 0;
        ans[0] = '0';
        ans[1] = '.';
        for(int i = 0 ; i < n ; i++){
            ans[2+i] = '0';
        }
        ans[n+2] = '\0';
    }else{
        e = point>not0?point-not0:point-not0+1;   // 指数
        ans[0] = '0';
        ans[1] = '.';
        for(int i = 0,j = not0 ; i < n ; i++,j++){
            if(j < len){
                if(str[j] != '.'){
                    ans[2+i] = str[j];
                }else
                    i--;
            }else
                ans[2+i] = '0';
        }
        ans[2+n] = '\0';
    }
}
int main(){
    freopen("in.txt","r",stdin);
    scanf("%d %s %s",&n,s1,s2);
    char ans1[maxn],ans2[maxn];
    int e1,e2;
    solve(s1,ans1,n,e1);
    solve(s2,ans2,n,e2);
    if(strcmp(ans1,ans2) == 0 && e1 == e2){
        printf("YES ");
        printf("%s*10^%d\n",ans1,e1);
    }else{
        printf("NO ");
        printf("%s*10^%d %s*10^%d\n",ans1,e1,ans2,e2);
    }

    return 0;
}

运行结果：