PAT Basic 1048 digital encryption (20 points)

This problem required to achieve a digital encryption method. First, encrypted with a fixed positive integer A, for any positive integer B, and the numbers on each of positions corresponding to 1-bit digital to A following operation: After an odd bit, the corresponding bit of the digital sum modulo 13 - - 10 used here representative of J, representative of Q 11, K 12 represents; on even bit, with the number B minus the number of a, if the result is negative, then add 10. Here a bit to make the first one.

Input formats:

Given in one row sequentially input A and B, no more than 100 are positive integers, separated by a space therebetween.

Output formats:

The output encrypted in a row.

Sample input:

1234567 368782971

Sample output:

3695Q8118

Author: CHEN, Yue

Unit: Zhejiang University

Time limit: 400 ms

Memory Limit: 64 MB

 

#include<iostream>
#include<stack>
using namespace std;
int main() {
    string s,s2;
    cin>>s>>s2;
    stack<int> sta1,sta2;
    stack<char> res;
    int tmp1,tmp2,tmpRes;
    for(int i=0;i<s.length();i++){
        sta1.push(s[i]-'0');
    }
    for(int i=0;i<s2.length();i++){
        sta2.push(s2[i]-'0');
    }
    bool isOdd=true;
    while(!sta1.empty()||!sta2.empty()){
        //取数值
        if(!sta1.empty()) {
            tmp1=sta1.top();
            sta1.pop();
        }else{
            tmp1=0;
        }
        if(!sta2.empty()) {
            tmp2=sta2.top();
            sta2.pop();
        }else{
            tmp2=0;
        }
        //奇偶处理数据
        if(isOdd){
            tmpRes=(tmp1+tmp2)%13;
            switch(tmpRes){
                case 10:res.push('J');break;
                case 11:res.push('Q');break;
                case 12:res.push('K');break;
                default:res.push(tmpRes+'0');break;
            }
        }else{
            tmpRes=tmp2-tmp1;
            if(tmpRes<0)tmpRes+=10;
            res.push(tmpRes+'0');
        }
        isOdd=!isOdd;
    }
    //output
    while(!res.empty()){
        cout<<res.top();
        res.pop();
    }
    system("pause");
    return 0;
}