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  • Uva 10014 Simple calculations

     Simple calculations 

    The Problem

    There is a sequence of n+2 elements a0, a1,…, an+1 (n <= 3000; -1000 <=  ai 1000). It is known that ai = (ai–1 + ai+1)/2 – ci   for each i=1, 2, ..., n. You are given a0, an+1, c1, ... , cn. 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 an+1 each having two digits after decimal point, and the next n lines contain numbers ci (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 an+1.

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

    Sample Input

    1
    
    1
    50.50
    25.50
    10.15
    

    Sample Output

    27.85


    #include<stdio.h>
    #include<string.h>
    
    int main()
    {
        int T, n, i;
        double first, final, sum, temp, ans;
        scanf("%d", &T);
        while(T--)
        {
            scanf("%d", &n);
            scanf("%lf", &first);
            scanf("%lf", &final);
            for(i=1,sum=0; i<=n; ++i)
            {
                scanf("%lf", &temp);
                sum += 2.0*(n+1-i)*temp;
            }
            
            ans = n*first+final-sum;
            printf("%.2lf\n", ans/(n+1));
            
            if(T != 0) printf("\n");
        }
        return 0;
    }

     

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  • 原文地址:https://www.cnblogs.com/liaoguifa/p/2948629.html
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