```cpp
#include <stdio.h>
int main()
{
char u[7],e[7];
int a = 1, b = 1, i;
scanf("%s%s", u, e);
for(i = 0; u[i] != NULL; i++)
a = a*(u[i]-'A'+1)%47;
for(i = 0; e[i] != NULL; i++)
b = b*(e[i]-'A'+1)%47;
if(a == b) printf("GO");
else printf("STAY");
return 0;
}
```
by 仇哥 @ 2016-09-19 23:26:44
那么就用c++的string啊
for(string::iterator i=a.begin();i!=a.en();i++)
by ghj1222 @ 2016-09-20 06:57:38
用'\0'也可以，直接u[i]做判断也可以。
by x_faraway_x @ 2016-09-20 12:44:17
膜各位大佬
by sslovelyq @ 2016-10-06 12:40:26
用str.length()行了。
by Wh_Xcjm @ 2016-10-06 15:29:16
敢问大神这个解法怎么错的
```cpp
#include <stdio.h>
#include <math.h>
#include <string.h>
int main()
{
char a[6],b[6];
int x=1,y=1,n;
scanf("%s",&a);
scanf("%s",&b);
for(n=0;n<6;n++)
{
if(a[n]!=0)x*=a[n]-64;
if(b[n]!=0)y*=b[n]-64;
}
if((x%47)==(y%47))printf("GO");
else printf("STAY");
return 0;
}
```
by IcaRuS @ 2016-10-25 20:45:19
是 不需要用strlen 用XXX.size()
by 特大号的黑音 @ 2016-11-04 20:19:34