HDU 1005:Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 141409    Accepted Submission(s): 34334

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

 

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#include<stdio.h>
int main()
{
    int a,b,n,z=1;
    int f[54]= {0,1,1};
    while(scanf("%d%d%d", &a, &b, &n)!= EOF)
    {
        if(a==0&&b==0&&n==0)break;
        for(int i = 3; i < 54; ++i)
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(i>5)
                if(f[i-1]==f[3]&&f[i]==f[4])
                {
                    z=i-4;
                    break;
                }
        }
        if(n>2)printf("%d
",f[(n-3)%z+3]);
        else printf("1
");
    }
    return 0;
}