PTA乙级 (1048 数字加密 (20分))

1048 数字加密 (20分)

https://pintia.cn/problem-sets/994805260223102976/problems/994805276438282240

第一次提交:

 错误原因:a的位数大于b时,b不足的位需要补0做运算!

 

第二次提交:

 代码:

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#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstdio>
using namespace std;
int main()
{
    int a[102]={0},b[102]={0};
    int i=1,j=1,k=1;
    char chr1[102],chr2[102];
    int len1,len2,len;
    cin>>chr1>>chr2;
    len1=strlen(chr1);
        len2=strlen(chr2);
    for(i=len1-1;i>=0;i--)
    {
        a[k]=chr1[i]-'0';
        k++;
    }
    for(i=len2-1;i>=0;i--)
    {
        b[j]=chr2[i]-'0';
        j++;
    }
    if(len1>len2) len=len1;
    else len=len2;
    for(i=1;i<=len;i++)
    {
        if(i%2==1) b[i]=(b[i]+a[i])%13; 
        else{
            if(b[i]<a[i]) b[i]=b[i]-a[i]+10;
            else b[i]=b[i]-a[i];
            }
    }
    for(i=len;i>=1;i--)
    {
        if(b[i]==10) cout<<"J";
        else if(b[i]==11) cout<<"Q";
        else if(b[i]==12) cout<<"K";
        else cout<<b[i];
    }
    return 0;
}
posted @   yyer  阅读(332)  评论(0编辑  收藏  举报
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