poj 2601 Simple calculations

最新推荐文章于 2020-01-28 16:56:14 发布

原创最新推荐文章于 2020-01-28 16:56:14 发布 · 899 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #算法 #编程 #poj #zoj

OJ编程专栏收录该内容

81 篇文章

订阅专栏

本文介绍了一种通过已知序列的前两个元素和中间元素之间的关系来计算序列中任意一个元素的方法，包括输入数据格式、解题思路以及实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Simple calculations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6559		Accepted: 3291

Description

There is a sequence of n+2 elements a₀, a₁, ..., a_n+1 (n <= 3000, -1000 <= a_i <=1000). It is known that a_i = (a_i-1 + a_i+1)/2 - c_i for each i=1, 2, ..., n.
You are given a₀, a_n+1, c₁, ... , c_n. Write a program which calculates a₁.

Input

The first line of an input contains an integer n. The next two lines consist of numbers a₀ and a_n+1 each having two digits after decimal point, and the next n lines contain numbers c_i (also with two digits after decimal point), one number per line.

Output

The output file should contain a₁ in the same format as a₀ and a_n+1.

Sample Input

Sample Output

27.85

解题思路：

大概过程：
a[0]+a[2]-2a[1]-2c[1]=0
a[1]+a[3]-2a[2]-2c[2]=0
……
a[n-1] + a[n+1] - 2a[n] - 2c[n] = 0
累加可得：
a[0]+a[n+1]-a[1]-a[n]-2c[1]-2c[2]-...-2c[n]=0
根据a[n-1]+a[n+1]-2a[n]-2c[n]=0  => a[n+1]-2c[n]-a[n]=a[n]+2c[n]-a[n-1]
化简：a[0]+a[n]-a[1]-a[n-1]-2c[1]-2c[2]-...-2c[n-1]=0
同理：a[0]+a[n-1]-a[1]-a[n-2]-2c[1]-2c[2]-...-2c[n-2]=0
      ……
      a[0]+a[2]-a[1]-a[1]-2c[1]=0
相加上面各式可得n*a[0]+a[n+1]-(n+1)*a[1]-2*n*c[1]-2*(n-1)*c[2]-...-2*c[n]=0
即a[1]=(n*a[0]+a[n+1]-2*n*c[1]-2*(n-1)*c[2]-...-2*c[n])/(n+1)

#include <iostream>
#include <iomanip>
using namespace std;
#define MAX 3005
int main(){
	int n;
	double a0,an;
	double c[MAX];
	while (cin>>n){
		cin>>a0>>an;
		double ans=0;
		for (int i=0;i<n;i++){
			cin>>c[i];
			ans+=2*(n-i)*c[i];
		}
		ans=(n*a0+an-ans)/(n+1);
		cout<<fixed<<setprecision(2)<<ans<<endl;
	}
	return 0;
}