poj 2231Moo Volume

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20699 Accepted: 6249

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.

解题思路：本题要求的就是任意两数之差的和。。。如果直接两层循环肯定会超时，这里可以利用动态规划的思想求解。

首先我们将这些数按从小到大排序，我们假定dp[i]为第i个数与前i-1个数差和，那么我们可以推导出递推公式:

dp[i] = dp[i-1]+(i-1)*(a[i] - a[i-1])。。最后将dp[i]都累加起来，记得还要乘以2，因为是双向的，dp[i]只是单向。。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 10010;int n;__int64 dp[maxn],a[maxn];int cmp(__int64 a,__int64 b){return a < b;}int main(){while(scanf("%d",&n)!=EOF){for(int i = 1; i <= n; i++)scanf("%I64d",&a[i]);sort(a+1,a+n+1,cmp);for(int i = 1; i <= n; i++){dp[i] = dp[i-1]+(i-1)*(a[i]-a[i-1]);}__int64 sum = 0;for(int i = 1; i <= n; i++)sum += dp[i];printf("%I64d\n",sum<<1);}return 0;}