PAT 乙级 1017

比较有技巧的一题。

#include<bits/stdc++.h> using namespace std; int main() { string a; int b; cin >> a >> b; int temp = 0; if(a.size() == 1) cout << (a[0] - '0') / b << " " << (a[0] - '0') % b; else if(a.size() > 1 && a[0] - '0' >= b) { for(int i = 0; i < a.size(); i ++) { temp = (temp * 10 + a[i] - '0'); cout << temp / b; temp = temp % b; } cout << " " << temp; } else if(a.size() > 1 && a[0] - '0' < b) { temp = a[0] - '0'; for(int i = 1; i < a.size(); i ++) { temp = temp * 10 + a[i] - '0'; //字符减0，得到数字 cout << temp / b; temp = temp % b; } cout << " " << temp; } return 0; }