PAT 乙级 1048 数字加密

Posted 2020-11-20 fdprocess

输入样例：
1234567 368782971
输出样例：
3695Q8118

思路很容易就能想到。注意，如果两个数字长度不相等，需要在短的那一个数字前面补零。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;

void encryption(vector<int> s1,vector<int> s2);

int main()
{
    vector<int> s1,s2;
    char c;
    c=getchar();
    while(c!=‘ ‘&&c!=‘
‘)
    {
        s1.push_back(c-‘0‘);
        c=getchar();
    }
    c=getchar();
    while(c!=‘ ‘&&c!=‘
‘)
    {
        s2.push_back(c-‘0‘);
        c=getchar();
    }
    //补全两个字符串的长度
    if(s1.size()>s2.size())
    {
        while(s1.size() != s2.size())
            s2.insert(s2.begin() ,0);
    }
    if(s1.size()<s2.size())
    {
        while(s1.size() != s2.size())
            s1.insert(s1.begin() ,0);
    }
    encryption(s1,s2);
    return 0;
}

void encryption(vector<int> s1,vector<int> s2)
{
    reverse(s1.begin(),s1.end());
    reverse(s2.begin(),s2.end());
    for(int i=0; i<s2.size(); i++)
    {
        if((i+1)%2==0)
        {
            s2[i]-=s1[i];
            if(s2[i]<0)
                s2[i]+=10;
        }
        else
            s2[i]=(s2[i]+s1[i])%13;
    }
    reverse(s2.begin(),s2.end());
    for(vector<int>::iterator itor=s2.begin(); itor!=s2.end() ; itor++)
    {
        switch(*itor)
        {
            case 10:
                cout<<‘J‘;
                break;
            case 11:
                cout<<‘Q‘;
                break;
            case 12:
                cout<<‘K‘;
                break;
            default:
                cout<<*itor;
        }
    }
}