The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
For each test case, print the value of f(n) on a single line.
2
5
ps:本按正常思路写，发现超时。。然后就输数据。。发现原来还有循环。。也就是说找到相对应的循环节就好
#include "stdio.h"
int main()
{
int a,b,n,f[100],i;
while(~scanf("%d%d%d",&a,&b,&n)&&a&&b&&n)
{
f[1]=1;f[2]=1;
for(i=3;i<100;i++)
{
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(f[i-1]==1&&f[i]==1)
break;
}
i-=2;
if(n%i!=0)
printf("%d\n",f[n%i]);
else
printf("%d\n",f[i]);
}
return 0;
}
