1636

比较简单的一道题目，统计00和01的个数之和为b（其实b也就是0的个数），统计00的个数为a，那么a/b就是rotate操作之后能够存活的概率，b/n(n代表的是可以装火药的位置个数)也就是shoot操作能够存活的概率，比较二者即可，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;int main(){string s;while (cin >> s){int n = s.size();int a = 0, b = 0;//b 代表的是0的个数  a 代表的是00的个数for (int i = 0; i < n; i++){if (s[i] == '0'){b++;if (s[(i + 1) % n] == '0') a++;}}if (a*n > b*b) cout << "SHOOT\n";else if (a*n < b*b) cout << "ROTATE\n";else cout << "EQUAL\n";}return 0;}