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  • POJ 2231 Moo Volume

    Moo Volume
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 18163   Accepted: 5347

    Description

    Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. 

    FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.

    Input

    * Line 1: N 

    * Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).

    Output

    There are five cows at locations 1, 5, 3, 2, and 4.

    Sample Input

    5
    1
    5
    3
    2
    4

    Sample Output

    40
    题目大意:  给定一串数字,求任意两个数字之差的和。
    #include <stdio.h>
    #include <iostream>
    #include <stdlib.h>
    #include <algorithm>
    using namespace std;
    
    int main()
    {
        int n;
        scanf("%d", &n);
        long long temp[10005] = {0};
        long long a[10005] = {0};
        long long sum = 0;
        for (int i = 1; i <= n; i++)
        {
            scanf("%lld", &a[i]);
        }
        sort(a + 1, a + n + 1);
        for (int i = 2; i <= n; i++)
        {
            temp[i] = temp[i - 1] + (i - 1) * (a[i] - a[i - 1]);
            sum += temp[i];
        }
        printf("%lld
    ", sum * 2);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lzmfywz/p/3191714.html
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