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

5 1 5 3 2 4

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 10100 int 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; }