UVa 10014 Simple calculations

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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

给跪了。。。推公式，其实早就推出来了，但是没有看题目要求要就输出a[1]，结果一直在化简。。。处于不知道在求什么的状态。。。

代码如下：

#include <stdio.h>
int main(void)
{int cases,n,i,j;double a,b,c[3002],sum;scanf("%d\n",&cases);while(cases--){scanf("%d",&n);scanf("%lf\n",&a);scanf("%lf",&b);for(i=1;i<=n;i++)scanf("%lf",&c[i]);sum=0.0;for(i=1;i<n+1;i++)for(j=1;j<=i;j++)sum+=c[j];sum*=2;a=(n*a+b-sum)/(n+1.0);printf("%4.2f\n",a);  if(cases)printf("\n");}return 0;
}

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