HDU_1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40967    Accepted Submission(s): 8785

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

找到规律，打表就可以了

#include <stdio.h>
int f[1009];
int main()
{
    int a, b, i, n, c;
    f[1] = 1;
    f[2] = 1;
    while(scanf("%d%d%d",&a,&b,&n) && (a || b || n))
    {
        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;
        }
        c=i-2;
        n=n%c;
        if(n == 0)
            printf("%d\n",f[c]);
        else
            printf("%d\n",f[n]);
    }
    return 0;
}