#include<stdio.h>
int main()
{
    int fun(int a,int b,int n);
    int a,b,n,N;
    while(scanf("%d%d%d",&a,&b,&n)!=EOF)
    {
        if(!(a==0&&b==0&&n==0))
        {
        N=fun(a,b,n);
        printf("%d\n",N);
        }
    }
    return 0;
} 
int fun(int a,int b,int n)
{
    int N;
    if(n>2)
    {
        N=(a*fun(a,b,n-1)+b*fun(a,b,n-2))%7;
        return N;
    }
    else
    return 1;
}

[此贴子已经被作者于2017-2-27 07:38编辑过]

一般是内存溢出。你这个可能是输入的数据过大，栈溢出了。

那怎么解决呢

怎么样改写程序，可以避免栈溢出呢

首先得提出题目要求

Problem Description
A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).
 

Input
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.
 

Output
For each test case, print the value of f(n) on a single line.
 

Sample Input
1 1 3
1 2 10
0 0 0
 

Sample Output
2
5

弃用递归，改用循环就可。

首先不应该用递归，你这递归中有太多的重复计算
其次，f(n)跟f(n-1)和f(n-2)相关，且f(n-1)和f(n-2)都小于7，所以f(n)的序列是循环的，最多七七四十九节长