一道关于字符串处理的题

题目要求：按要求分解字符串，输入两个数M，N；M代表输入的M串字符串，N代表输出的每串字符串的位数，不够补0。例如：输入2,8， “abc” ，“123456789”，则输出为“abc00000”,“12345678“,”90000000”

#include<iostream>#include<iomanip>#include <stdio.h>#include <string.h>#include <list>#include <vector>#include <cstdio>#include <stack>#include <queue>#include <sstream>using namespace std;int main(){int M, N;cin >> M >> N;vector<string> strs;for (int i = 0; i < M; i++){string s;cin>>s;strs.push_back(s);}for (int i = 0; i < strs.size(); i++){string s = strs[i];int slen = s.length();if (slen <= N){for (int j = 0; j < slen; j++){cout << s[j];}for (int k = 0; k < N - slen; k++){cout << "0";}cout << endl;}else{int num_sy = slen%N;int num_zero = N - num_sy;for (int x = 0; x < num_zero; x++){s.append("0");}for (int y = 0; y < s.length(); y++){if (y%N == 0){cout << endl;}cout << s[y];}}}return 0;}