杭电 1005 Number Sequence

来源：互联网发布：深圳前海共享网络编辑：程序博客网时间：2024/06/06 12:48

Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45754 Accepted Submission(s): 10082
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

代码部分：

<----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------->

#include"stdio.h"
int main()
{
    int A,B,n,i,f[200];
    while(scanf("%d %d %d",&A,&B,&n)!=EOF)
    {

        if(A==0&&B==0&n==0)break;
        if(n<=1)
            {
                printf   ("1\n");
                continue;
            }
        f[1]=1,f[2]=1;
        A%=7,B%=7;
        for(i=3;i<=52;i++)
            {
                f[i]=(A*f[i-1]+B*f[i-2])%7;
                if(f[i-1]==1 && f[i]==1) break;
            }
        i=i-2;
        n%=i;
        f[0]=f[i];
        printf("%d\n",f[n]);
    }
    return 0;
}

0 0