Number Sequence

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

 

#include <stdio.h>   //N要满足条件的足够大，至于为什么是201要慢慢增大N的数值进行尝试
#define N 201

int main(){
    int A;
    int B;
    long n;
    int i;
    int f[N];
    int flag;

    f[1]=1;
    f[2]=1;

    while(1){
        scanf("%d%d%ld",&A,&B,&n);

        if(A==0 && B==0 && n==0)
            break;

        flag=0;
        for(i=3;i<N;i++){   //两种情况，一种是以1 1开头进行循环的数列，另一种是1 1 0 0 0...
            f[i]=(f[i-1]*A+f[i-2]*B)%7;
            
            if(f[i]==1 && f[i-1]==1)
                break;

            if(f[i]==0 && f[i-1]==0){
                flag=1;
                break;
            }
        }

        if(flag==1){
            if(n==1 || n==2)
                printf("1
");

            else
                printf("0
");
            continue;
        }

        i-=2;   //i为周期
        n%=i;   //n为周期内的某个数
        
        if(n==0)  //f[0]=f[i]
            n=i;

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

    return 0;
}