Number Sequence（存在循环规律）

最新推荐文章于 2020-07-15 21:35:54 发布

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最新推荐文章于 2020-07-15 21:35:54 发布

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分类专栏： ————数学问题———— 水题文章标签： acm

本文链接：https://blog.youkuaiyun.com/qaz135135135/article/details/51997964

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本文介绍了一个特定数列的计算方法，该数列定义为f(1)=1, f(2)=1, f(n)=(A*f(n-1)+B*f(n-2))mod7。文章提供了一种高效算法来计算数列中任意项的值，通过发现数列每48项重复一次的规律避免了超时问题。

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1803 Accepted Submission(s): 688

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
每48个一组循环，直接写会超时；
#include<stdio.h>
#include<string.h>
int main()
{
    int a,b,n;
    while(scanf("%d%d%d",&a,&b,&n)==3)
    {
        if(!a&&!b&&!n)
            break;
        int arr[48];
        arr[1]=1;
        arr[2]=1;
        for(int i=3;i<48;i++)
            arr[i]=((a*arr[i-1]+b*arr[i-2])%7);
        printf("%d\n",arr[n%48]);
    }
    return 0;
}