Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 149022 Accepted Submission(s): 36261

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

不知道为什么把T=i-2放到循环里面直接break会一直WA,re,,严重怀疑后台数据。。找循环节就行了..因为f[i-1] f[i-2]都属于[0,6]所以所有的排列不会超过49种，，在49次里面总会出现循环。。水题卡了我好久。。

#include <stdio.h>

int main()

{

    int A,B,n,f[];

    while(scanf("%d%d%d",&A,&B,&n)!=EOF,A||B||n)

    {

        f[]=,f[]=;

        int i;

        for(i=; i<; i++)

        {

            f[i] = (A*f[i-]+B*f[i-])%;

            if(f[i]==&&f[i-]==)

            {

                break;

            }

        }

        int T=i-;

        f[]=f[T];

        printf("%d\n",f[n%T]);

    }

    return ;

}

巴特西

hdu 1005(找循环节)

Number Sequence

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