杭电acm第1005题Number Sequence

最新推荐文章于 2021-02-17 03:11:11 发布

Paul-Yang

最新推荐文章于 2021-02-17 03:11:11 发布

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文章标签：杭电 acm 杭电acm第1005题Number Se

本文链接：https://blog.youkuaiyun.com/shouwang_tomorrow/article/details/45727261

本文深入探讨了NumberSequence问题的解决策略，包括输入解析、周期寻找和输出生成。通过实例代码展示了如何利用循环和模运算高效求解给定序列的特定项。

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 123152 Accepted Submission(s): 29915

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

注：此题为：杭电acm第1005题Number Sequence

已AC代码：

#include<stdio.h>
int main()
{
    int A,B,i;
    long n;
    int f[201];
    f[1]=f[2]=1;
    while(scanf("%d %d %ld",&A,&B,&n))
    {
        if(A==0&&B==0&&n==0) break;
        int cnt=0;
        for(i=3;i<=200;i++)//寻找周期 
        {
            f[i]=(A*f[i-1]+B*f[i-2])%7;
            if(f[i]==1&&f[i-1]==1)break;
            if(f[i]==0&&f[i-1]==0)
			{
			    cnt=1;break;
			}//如果A=7，B=7则后面都为0了 
        }
        if(cnt){printf("0\n");continue;}
        if(i>n){printf("%d\n",f[n]);continue;}
        i-=2;//i为周期
        n%=i;
        if(n==0)n=i;
        printf("%d\n",f[n]);    
    } 
    return 0;       
}