poj2231 2010.8.1

Moo Volume

Time Limit: 1000MS  Memory Limit: 65536K

Total Submissions: 11299  Accepted: 3177

 

Description

 

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

 

FJ's N cows (1 <= N <= 10,000) allgraze at various locations on a long one-dimensional pasture. The cows are verychatty animals. Every pair of cows simultaneously carries on a conversation (soevery cow is simultaneously MOOing at all of the N-1 other cows). When cow iMOOs at cow j, the volume of this MOO must be equal to the distance between iand j, in order for j to be able to hear the MOO at all. Please help FJ computethe total volume of sound being generated by all N*(N-1) simultaneous MOOingsessions.

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 5contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is(10+10+6+7+7) = 40.

Source

 

USACO 2005 January Silver

 

分析：

水题，但是。。。

刚开始觉得太水了，暴力就能过。。。但是暴力是不行的。。后来搜了解题报告，发现，是一道递推+数学的题。。。。

         1.先把牛的坐标从小到大排序；

         2.牛i 和牛i-1  之间的这段距离，被用了的次数=这个距离的左边的牛数*这个距离右边的牛数；

         3.把这些都加起来，*2，就是结果了。

很强，我都米想到。。。。一定要抓住本质。。。

 

Source Code

 

Problem: 2231  User: creamxcream

Memory: 216K  Time: 16MS

Language: C++  Result: Accepted

 

Source Code 

#include <iostream>#include <algorithm>using namespace std;#define MAXN 10010__int64 position[MAXN];int n;bool cmp(__int64 a,__int64 b){return (a<b);}int main(){scanf("%d",&n);memset(position,0,sizeof(position));for(int i=1;i<=n;i++)scanf("%I64d",&position[i]);sort(position+1,position+n+1,cmp);__int64 ans=0;for(int i=2;i<=n;i++)ans+=(position[i]-position[i-1])*(i-1)*(n-i+1);printf("%I64d\n",ans*2);return 0;}