hdu 1005(找循环节)

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[55];
    while(scanf("%d%d%d",&A,&B,&n)!=EOF,A||B||n)
    {
        f[1]=1,f[2]=1;
        int i;
        for(i=3; i<50; i++)
        {
            f[i] = (A*f[i-1]+B*f[i-2])%7;
            if(f[i]==1&&f[i-1]==1)
            {
                break;
            }
        }
        int T=i-2;
        f[0]=f[T];
        printf("%d
",f[n%T]);
    }
    return 0;
}