This problem requires the implementation of a digital encryption method. Firstly, a positive integer A is fixed for encryption, and every positive integer B is computed as follows: for odd digits, the digits of corresponding digits are added to the digits of A, and then 13 is redundant, where J stands for 10, Q stands for 11, K stands for 12; for even digits, the digits of B are subtracted from the digits of A, and if the result is negative, 10 is added. Let's make it number one here.
Input format:
Inputs in a row give A and B in turn, both positive integers of no more than 100 bits, separated by spaces.
Output format:
Output the encrypted result in one line.
Input sample:
1234567 368782971
Output sample:
3695Q8118
Author: CHEN, Yue
SETTING: 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()){ //Take numerical value if(!sta1.empty()) { tmp1=sta1.top(); sta1.pop(); }else{ tmp1=0; } if(!sta2.empty()) { tmp2=sta2.top(); sta2.pop(); }else{ tmp2=0; } //Parity Processing Data 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; }