hdu 2817 A sequence of numbers

最新推荐文章于 2020-07-07 21:38:00 发布

原创最新推荐文章于 2020-07-07 21:38:00 发布 · 381 阅读

0 ·

CC 4.0 BY-SA版权

hdu oj 同时被 2 个专栏收录

106 篇文章

订阅专栏

数学

22 篇文章

订阅专栏

本文探讨了如何在给定三个数列项的情况下，识别数列是等差还是等比，并通过中国剩余定理或快速幂取余法计算第k项值，同时提供了求解过程的详细步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A sequence of numbers

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 47 Accepted Submission(s) : 21

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

Xinlv wrote some sequences on the paper a long time ago, they might be arithmetic or geometric sequences. The numbers are not very clear now, and only the first three numbers of each sequence are recognizable. Xinlv wants to know some numbers in these sequences, and he needs your help.

Input

The first line contains an integer N, indicting that there are N sequences. Each of the following N lines contain four integers. The first three indicating the first three numbers of the sequence, and the last one is K, indicating that we want to know the K-th numbers of the sequence.

You can assume 0 < K <= 10^9, and the other three numbers are in the range [0, 2^63). All the numbers of the sequences are integers. And the sequences are non-decreasing.

Output

Output one line for each test case, that is, the K-th number module (%) 200907.

Sample Input

2
1 2 3 5
1 2 4 5

Sample Output

5
16

Source

题目大意：给出一个数列，要么是等差，要么是等比，现在给出前三项，打印第k项

思路：对于等差数列，中国剩余定理即可，对于等比数列，采用快速幂取余即可对付：

#include<stdio.h>
#include<string.h>
#include<math.h>
__int64 quick(__int64 m,__int64 n,__int64 k)
{
	__int64 b=1;
	m%=k;
	while(n>0)
	{
		if(n&1) //n是奇数
			b=(b*m)%k;
		n=n>>1;
		m=(m*m)%k; 
	}
	return b;
}
int main()
{
	int n;
	__int64 a,b,c,k;
	__int64 temp1,temp2;
//	double temp3,temp4;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k);
		temp1=b-a;
		temp2=c-b;
		if(temp1==temp2)
		{
			printf("%d\n",(a%200907+(k-1)%200907*temp1%200907)%200907);
			continue;
		}
	
	    else
	    {
	    	temp2=b/a;
	    	printf("%d\n",(a*quick(temp2,k-1,200907)%200907));
	    }
		
		
	}
return 0;
}