POJ2231奶牛叫

Moo Volume

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22151 Accepted: 6712

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

515324

Sample Output

40

Hint

INPUT DETAILS: 

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

OUTPUT DETAILS: 

Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.

    看了这个题目，开始不明白什么意思，后来看了HINT的解释才明白。简单点，就是求n个数两两之差的和，第一想到的是暴力法解，但又怕10000*10000=10^8会超时，压界（希望北大的服务器厉害）。下面是我的代码

#include <stdio.h>#include <stdlib.h>#define MAX 10100int main (){    int n;    long long num[MAX];    long long sum = 0;    scanf ("%d", &n);    for (int i = 1; i <= n; i++)        scanf ("%lld", &num[i]);    for (int i = 1; i < n; i++)        for (int j = i + 1; j <= n; j++)            sum += abs(num[i] - num[j]);    printf ("%lld\n", sum * 2);    return 0;}

    就这样AC了，就这样AC了，就这样AC了！！但老师把这一题归在了递推题里面，我就尝试着用递推写，草稿纸浪费了不少，规律矩阵我能找出来，但是用代码实现起来很复杂。我就去大佬的CSDN看他们的解题思路，结果发现他们的规律矩阵跟我的完全不同，看了几天我也不知道这种思路，无奈只好拿别人的规律来了

#include <stdio.h>#include <stdlib.h>int cmp(const void* a,const void* b){    return *(int*)a-*(int*)b;}int main(){    long long ans = 0;    int n,m,arr[10010];    scanf("%d",&n);    for(int i = 0; i < n; ++i)        scanf("%d",&arr[i]);    qsort(arr,n,sizeof(int),cmp);    for(int i = 0; i < n; ++i)        ans += (n-1-i) * (arr[n-1-i] - arr[i]);    printf("%lld\n",2*ans);    return 0;}