玩个烂梗

#include<bits/stdc++.h>
using namespace std;
int main(){
    char c[7],g[7];
    int ge=1,neetimay=1;
    scanf("%s%s",c,g);
    for(int i=0;i<6;i++) if(c[i]!='\0') ge*=(c[i]-'A'+1);
    ge%=47;
    for(int i=0;i<6;i++) if(g[i]!='\0') neetimay*=(g[i]-'A'+1);
    neetimay%=47;
    if(ge==neetimay) printf("GO");
    else printf("STAY");
    return 0;
}

using namespace std; int main(){ char c[7],g[7]; int ge=1,neetimay=1; scanf("%s%s",c,g); for(int i=0;i<6;i++) if(c[i]!='\0') ge*=(c[i]-'A'+1); ge%=47; for(int i=0;i<6;i++) if(g[i]!='\0') neetimay*=(g[i]-'A'+1); neetimay%=47; if(ge==neetimay) printf("GO"); else printf("STAY"); return 0; }

代码都写在下面了（含注释）

#include <bits/stdc++.h>
using namespace std;
long long len1,len2,ans1=1,ans2=1,x; //一定要用longlong！用int也许会溢出。
string s1,s2;
int main() {
	cin>>s1>>s2;
	len1=s1.size(),len2=s2.size();//长度赋值。
	for (int i=0;i<len1;i++) {
		x=s1[i]-'A'+1;//计算当前字符的对应数字。
		ans1=ans1*x;//乘起来。
	}
	for (int i=0;i<len2;i++) {
		x=s2[i]-'A'+1;
		ans2=ans2*x;
        //同理
	}
	if (ans1%47==ans2%47) cout<<"GO";//判断是否可以去/不去。
	else cout<<"STAY";
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main()
{
    string ufo;
    string usaco;
    cin>>ufo;
    cin>>usaco;
    int sum = 1;
    int sum2 = 1;
    for (int i = 0;i <= ufo.size()-1;i++)
        sum *= ufo[i] - 64;
    for (int i = 0;i <= usaco.size()-1;i++)
        sum2 *= usaco[i] - 64;
    if (sum % 47 == sum2 % 47)
        cout<<"GO"<<endl;
    else
        cout<<"STAY"<<endl;
    return 0;
}

上代码（说明在代码里）：

#include<iostream>
#include<string>
using namespace std;
string s1,s2;
const unsigned int to_num[128]={255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255};//打个表，不用每次字母转数字都要算
unsigned int hashes(string s){//用于算一个字符串转数字后mod47的值
    unsigned int mul=1;
    for(unsigned int i=0,len=s.length();i<len;i++){//遍历字符串
        mul*=to_num[s[i]];
        mul%=47;//过程中就取余，防溢出
    }
    return mul;
}
int main(){
    cin >> s1 >> s2;
    if(hashes(s1)==hashes(s2)){//如果它们mod47的值相等
        cout << "GO";
    } else{
        cout << "STAY";
    }
    return 0;
}