bzoj1679[Usaco2005 Jan]Moo Volume 牛的呼声
2014-06-15 23:43
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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 variouslocations 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
51
5
3
2
4
INPUT DETAILS:
There are five cows at locations 1, 5, 3, 2, and 4.
Sample Output
40OUTPUT 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); }
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