杭电1005--Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 125495    Accepted Submission(s): 30510

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 num[55] ;   //数组开的要大于 i 最大值 ；       
int main()
{ 
    int a,b,c,i,temp ;
    num[1] = 1 ; num[2] = 1 ;
    while(~scanf("%d %d %d",&a, &b, &c) , (a || b || c) )
    {

        for(i=3; i<=49; i++)
        {
            num[i] = (a*num[i-1] + b*num[i-2]) %7 ;
            if(num[i] == num[i-1] && num[i] == 1 )
            break ;
        }
        temp = i-2 ;
        c = c%temp ;
        if(c == 0)
            printf("%d
",num[temp]) ;  //  一直理解错，取余后得到0为循环最后一个，不是第一个；
        else
            printf("%d
", num[c]) ;
    }
    return 0 ;
}