10014 - Simple calculations

最新推荐文章于 2017-09-06 15:16:57 发布

原创最新推荐文章于 2017-09-06 15:16:57 发布 · 667 阅读

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本文详细阐述了解决涉及复杂序列计算的问题，通过提供输入样例和解析步骤，展示了如何通过迭代公式计算序列中的特定项。重点在于理解序列的生成规则，并运用给定的初始条件和中间变量（如a0, an+1 和 ci）来推导目标值a1。通过实例演示，读者可以掌握解决类似问题的方法。

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Simple calculations

The Problem

There is a sequence of n+2 elements a₀, a₁,…, a_n+1(n <= 3000; -1000 <= a_i 1000). It is known thata_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

这题就是通过以下的方法找规律
n=1 : a2=2*a1+2*c1-a0 ；
n=2 : a3=3*a1+4*c1+2*c2-2*a0 ;
n=3 : a4=4*a1+6*c1+4*c2+2*c3-3*a0;
.
.
.
n=n :A(n+1)=(n+1)*a1+2*n*c1+(2*n-2)*c2+(2*n-4)*c3+....+2*Cn-n*a0;
这样就只有一个未知量a1了，就可以得出结果了

#include <stdio.h>
#include <string.h>
double a;
int main()
{
    int i,j,n,m,t;
    double s,s1;
    double x,y;
    scanf("%d",&m);
    while(m--)
    {
        s=0;
        s1=0;
        scanf("%d",&n);
        scanf("%lf %lf",&x,&y);
        t=2*n;
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&a);
            s1+=t*a;
            t=t-2;
        }
        s=s+y+n*x-s1;
        s=s/(n+1);
        printf("%.2lf\n",s);
        if(m)
        {
            printf("\n");
        }
    }
    return 0;
}