HDU1005

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

#include<stdio.h>
int f(int A,int B,int n)
{
    if(n==1||n==2)
        return 1;
    else
       return (A*f(A,B,n-1)+B*f(A,B,n-2))%7;

}
int main()
{
    int a,b,n;
    while(~scanf("%d%d%d",&a,&b,&n)&&n&&a&&b)
    {
        printf("%d\n",f(a,b,n));
    }
    return 0;
}

这个代码提交会出现超时，如果用f(a,b,n)会出现栈溢出，考虑f(n)=Af(n-1)+Bf(n-2)
由于f(n)是由前两个数字组合产生，那么只要有两个数字组合相同的情况发生就一定一会产生循环！
两个数字的组合的最大可能值为7（0–6）x7（0—6）=49，因此只要在调用迭代方法中限制n在0~48就可以了
优化如下：

#include<stdio.h>
int f(int A,int B,int n)
{
    if(n==1||n==2)
        return 1;
    else
       return (A*f(A,B,n-1)+B*f(A,B,n-2))%7;

}
int main()
{
    int a,b,n;
    while(~scanf("%d%d%d",&a,&b,&n)&&n&&a&&b)
    {
        printf("%d\n",f(a,b,n%49));
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
int s[50];
int main()
{
    int a,b,n,i;
    while(scanf("%d%d%d",&a,&b,&n),a || b || n)
    {
        int i;
        s[0]=s[1]=1;
        for(i = 2; i<50; i++)
        {
            s[i] = (a*s[i-1]+b*s[i-2])%7;
            if(s[i] ==1 && s[i-1] == 1)
            {
               break;
            }
        }
        n = n%(i-1);
        if(n == 0)
            printf("%d\n",s[i-2]);
        else
            printf("%d\n",s[n-1]);

    }
    return 0;
}