HDU1005

Number Sequence  HDU-1005

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 162730    Accepted Submission(s): 40016

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

 

 

发这道题主要是祭奠一下我第一次遇到的超内存MLE。。。。。第一想法这么简单，暴力递归。。然后MLE狠狠打脸。。

后来再认真读题，发现还是有规律的。

如果f(n - 1)和f(n)都为1的话，就是一个循环。

#include <iostream>
#include <cstdio>
using namespace std;
int f[51];
int main()
{
    int a, b;
    long long n;
    while (scanf("%d %d %lld", &a, &b, &n) == 3)
    {
        if (a == 0 && b == 0 && n == 0)
            break;
        f[1] = f[2] = 1;
        int i;
        for (i = 3; i < 51; i++)
        {
            f[i] = (a * f[i - 1] + b * f[i - 2]) % 7;
            if (f[i] == 1 && f[i - 1] == 1)   //找到循环因子 i  
            {
                break;
            }
        }
        n = n % (i - 2);
        if (n == 0)   //刚好经过一个循环 
            printf("%d
", f[i - 2]);   
        else
            printf("%d
", f[n]);
    }
    return 0;
}

这题真的是好烦。。。。。。。