UVA Simple calculations （数学推导）

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The Problem

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.

The Input

The first line is the number of test cases, followed by a blank line.

For each test case, the first line of an input file 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.

Each test case will be separated by a single line.

The Output

For each test case, the output file should contain a₁ in the same format as a₀ and a_n+1.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

1150.5025.5010.15

Sample Output

27.85

题意：给出a[0],和a[n+1]，以及c[1]～ｃ[n]的值，有公式

a[i] = (a[i-1] + a[i+1])/2 - c[i];

然后求a[1]的值。

根据公式推导:

a[1] = n*a[0] + a[n+1] -2*n*c[1] - 2*(n-1)*c[2] --->>> - 2*1*c[n];

a[1]就可以了.

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

using namespace std;

double c[100010];

int main()
{
    int T;
    double a0,an;
    scanf("%d",&T);
    while(T--)
    {
        double n;
        scanf("%lf",&n);
        scanf("%lf",&a0);
        scanf("%lf",&an);
        for(int i=1;i<=n;i++)
        {
            scanf("%lf",&c[i]);
        }
        double sum = 0;
        sum = n*a0 + an;
        for(int i=1;i<=n;i++)
        {
            sum = sum - 2.0 *(n-i+1) * c[i];
        }
        sum = sum / (n+1);
        printf("%.2lf\n",sum);
        if(T!=0)
            cout<<endl;
    }
    return 0;
}

UVA Simple calculations （数学推导）

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原文地址：http://blog.csdn.net/yeguxin/article/details/42836561