hdu 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147906 Accepted Submission(s): 35942

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<iostream>
#include<stdio.h>
using namespace std;
int dp[1500];
int main()
{
    int a,b,n;

    //cout<<maxx<<endl;

    while(~scanf("%d%d%d",&a,&b,&n)&&!(a==b&&b==n&&a==0))
    {
        int i;
       dp[0]= dp[1]=dp[2]=1;
        if(n<=2)
            printf("%d
",dp[n]);
        else
        {
            for(i=3; i<1500; i++)
            {
                dp[i]=(dp[i-1]*a+dp[i-2]*b)%7;
                if(dp[i-2]==1&&dp[i-1]==1&&i!=3)
                    break;
            }
            if(n%(i-3))
            printf("%d
",dp[n%(i-3)]);
            else printf("%d
",dp[i-3]);
        }

    }
    return 0;
}

View Code

这道题的标准做法是用矩阵快速幂。但是一个数不断的mod一个数，这个结果一定是一个周

期函数。所以可以通过找规律得到结果。是一个比较取巧的解法。