zhber
有好多做过的题没写下来,如果我还能记得就补吧

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.

    约翰的邻居鲍勃控告约翰家的牛们太会叫.
    约翰的N(1≤N≤10000)只牛在一维的草场上的不同地点吃着草.她们都是些爱说闲话的奶牛,每一只同时与其他N-1只牛聊着天.一个对话的进行,需要两只牛都按照和她们间距离等大的音量吼叫,因此草场上存在着N(N-1)/2个声音.  请计算这些音量的和.

Input

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

    第1行输入N,接下来输入N个整数,表示一只牛所在的位置.

Output

* Line 1: A single integer, the total volume of all the MOOs.

    一个整数,表示总音量.

Sample Input

5
1
5
3
2
4

INPUT DETAILS:

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

Sample Output

40

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.

我能说这是差分序列吗……

排序完搞出两两之间的距离,然后它被计算的次数是i*(n-i)

这不是tyvj原题吗

#include<cstdio>
#include<algorithm>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,a[10001],s[10001];
long long ans;
int main()
{
	n=read();
	for (int i=1;i<=n;i++)a[i]=read();
	sort(a+1,a+n+1);
	for (int i=1;i<n;i++)s[i]=a[i+1]-a[i];
	for (int i=1;i<n;i++)
	{
	  ans+=(long long) i*(n-i)*s[i];
	}
	ans*=2;
	printf("%lld",ans);
}


posted on 2014-06-15 23:43  zhber  阅读(174)  评论(0编辑  收藏  举报