这道题太水了,不讲了,直接上代码。
不过对于初学字符串的选手是道好题。
#include <iostream> using namespace std; int main() { string s; cin >> s; int a = 1, b = 1; for (int i = 0; i < s.length(); i++) a = (a * (s[i] - ‘A‘ + 1)) % 47; cin >> s; for (int i = 0; i < s.length(); i++) b = (b * (s[i] - ‘A‘ + 1)) % 47; cout << (a == b ? "GO" : "STAY"); return 0; }
持续更新
原文:https://www.cnblogs.com/zcr-blog/p/12627626.html