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[];
int main()
{
    int a, b;
    long long n;
    while (scanf("%d %d %lld", &a, &b, &n) == )
    {
        if (a ==  && b ==  && n == )
            break;
        f[] = f[] = ;
        int i;
        for (i = ; i < ; i++)
        {
            f[i] = (a * f[i - ] + b * f[i - ]) % ;
            if (f[i] ==  && f[i - ] == )   //找到循环因子 i
            {
                break;
            }
        }
        n = n % (i - );
        if (n == )   //刚好经过一个循环
            printf("%d\n", f[i - ]);
        else
            printf("%d\n", f[n]);
    }
    return ;
}

这题真的是好烦。。。。。。。