1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int f[4000];
 6     int a, b, n;
 7     while (cin >> a >> b >> n, a != 0 && b != 0 && n != 0)
 8     {
 9         int i;
10         f[1] = 1, f[2] = 1;
11         for (i = 3; i < 500; i++)
12         {
13             f[i] = (a*f[i - 1] + b * f[i - 2]) % 7;
14             if (f[i] == 1 && f[i - 1] == 1) break;
15          }
16         i -= 2;
17         if (i > n)
18         {
19             cout << f[n] << endl;
20             continue;
21         }
22         n = n % i;
23         if (n == 0) n = i;
24         cout << f[n] << endl;
25     }
26     return 0;
27 }

大佬代码：http://www.cnblogs.com/kuangbin/archive/2011/07/26/2117381.html

#include<stdio.h>
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int A,B,i;
    long n;
    int f[201];
    f[1]=f[2]=1;
    while(scanf("%d %d %ld",&A,&B,&n))
    {
        if(A==0&&B==0&&n==0) break;
        int cnt=0;
        for(i=3;i<=200;i++)//打表找到周期 
        {
            f[i]=(A*f[i-1]+B*f[i-2])%7;
            if(f[i]==1&&f[i-1]==1)break;
            if(f[i]==0&&f[i-1]==0){cnt=1;break;}//这里有个小陷阱，如果A=7，B=7则后面都为0了 
        }
        if(cnt){printf("0\n");continue;}
        if(i>n){printf("%d\n",f[n]);continue;}
        i-=2;//i为周期
        n%=i;
        if(n==0)n=i;
        printf("%d\n",f[n]);    
    } 
    return 0;   
    
}