HDU OJ 1005 Number Sequence

最新推荐文章于 2020-04-17 15:02:57 发布

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最新推荐文章于 2020-04-17 15:02:57 发布

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178988 Accepted Submission(s): 44485

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

Sample Output

2
5

Author

CHEN, Shunbao

Source

ZJCPC2004

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直接用递归或者循环肯定超时，爆内存。因为 1 <= n <= 100,000,000

所以关键在于 mod 7.

要找其循环周期。

如果是一个f()的话循环周期是7

现在是两个f()所以循环周期是49；

代码：

#include <iostream>  
using namespace std;  
int arr[50];  
int main()  
{  
    int n,a,b;  
    arr[1]=arr[2]=1;  
    while(cin>>a>>b>>n)  
    {  
        if(n==0) break;  
        for(int i=3; i<=50; i++)  
        {  
            arr[i]=(a*arr[i-1]+b*arr[i-2])%7;  
        }  
        cout<<arr[n%49]<<endl;  
    }  
    return 0;  
}

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