Number Sequence（http://acm.hdu.edu.cn/showproblem.php?pid=1005）

最新推荐文章于 2021-08-05 20:57:04 发布

LR1741670057

最新推荐文章于 2021-08-05 20:57:04 发布

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本文介绍了一个关于数列计算的问题，通过寻找数列的周期来高效计算特定项的值。利用模运算解决大规模数值计算问题，并提供了一段C语言代码实现。

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 129001 Accepted Submission(s): 31388

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

#include<stdio.h>
#include<string.h>
#define max 100000
int main()
{
    int a,b,n,i,j,x[max];
    while(scanf("%d%d%d",&a,&b,&n)&&(a||b||n))
    {
        x[1]=1;
        x[2]=1;
        for(i=3;i<=1001;++i)//找周期
        {
            x[i]=((a*x[i-1]+b*x[i-2])%7);
            if(x[i]==1&&x[i-1]==1)//周期的条件
            break;
        }
        int t=i-2;//确定周期
        int c=n%t;//判断
        if(c==0)
        printf("%d\n",x[t]);
        else
        printf("%d\n",x[c]);
    }
    return 0;
}