HDOJ 题目2187 A sequence of numbers（快速幂水）

最新推荐文章于 2021-01-22 21:54:59 发布

Jogging_Clown

最新推荐文章于 2021-01-22 21:54:59 发布

阅读量522

点赞数

CC 4.0 BY-SA版权

分类专栏：数学

本文链接：https://blog.youkuaiyun.com/yu_ch_sh/article/details/43496061

数学专栏收录该内容

164 篇文章

订阅专栏

本文介绍如何通过给定的前三个数字来判断序列是等比还是等差，并计算第K项的值，结果需对200907取模。

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: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3750 Accepted Submission(s): 1150

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

Sample Output

  
   5
16

Source

2009 Multi-University Training Contest 1 - Host by TJU

Recommend

gaojie | We have carefully selected several similar problems for you: 2819 2825 2822 2821 2820

无聊，切道氺题压压惊。

题目大意：判断是等比还是等差，并求出第n项，等差直接算，等比快速幂算

ac代码

#include<stdio.h>
#include<string.h>
#define mod 200907
__int64 qpow(__int64 a,__int64 b)
{
	__int64 ans=1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%mod;
		a=(a*a)%mod;
		b/=2;
	}
	return ans;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		__int64 a,b,c,d;
		scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&d);
		if(c-b==b-a)
		{
			printf("%I64d\n",(a+((d-1)*(c-b))%mod)%mod);
		}
		else
		{
			printf("%I64d\n",(a*(qpow(b/a,d-1))%mod)%mod);
		}
	}
}