杭电oj~1005 简单的找周期

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 153132    Accepted Submission(s): 37335

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

 

#include<stdio.h>
int fun(int,int,int );
int main()
{
    int a,b;
    long long int n;
while(scanf("%d %d %lld",&a,&b,&n)!=EOF&&(a||b||n))
{
    int arr[50]={1,1};
    for(int i=2;i<=48;i++)
    arr[i]=(a*arr[i-1]+b*arr[i-2])%7;
    if(n%49==0) printf("%d",arr[48]);
    else
        printf("%d
",arr[n%49-1]);
}
return 0;

}

　　

周期是7*7==49~yeah！

周期递归做会超时~毕竟n的规模是10^8,所以时间复杂度必须是O(n)!!!!!