HDOJ 5419 Victor and Toys 树状数组

分母是一定的C(m,3) 树状数组求每一个数能够在那些段中出现,若x出如今了s段中,分子加上w[x]*C(s,3)

Victor and Toys

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 331    Accepted Submission(s): 118

Problem Description

Victor has n toys, numbered from 1 to n. The beauty of the i-th toy is wi.

Victor has a sense of math and he generates m intervals, the i-th interval is [li,ri]. He randomly picks 3 numbers i,j,k(1≤i<j<k≤m), and selects all of the toys whose number are no less than max(li,lj,lk) and no larger than min(ri,rj,rk). Now he wants to know the expected sum of beauty of the selected toys, can you help him?

 

Input

The first line of the input contains an integer T, denoting the number of test cases.

In every test case, there are two integers n and m in the first line, denoting the number of the toys and intervals.

The second line contains n integers, the i-th integer wi denotes that the beauty of the i-th toy.

Then there are m lines, the i-th line contains two integers li and ri.

1≤T≤10.

1≤n,m≤50000.

1≤wi≤5.

1≤li≤ri≤n.

 

Output

Your program should print T lines : the i-th of these denotes the answer of the i-th case.

If the answer is an integer, just print a single interger, otherwise print an irreducible fraction like p/q.

 

Sample Input

1
3 4
1 1 5
2 3
1 3
3 3
1 1

 

Sample Output

5/4

 

Source

BestCoder Round #52 (div.2)

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月23日 星期日 14时23分47秒
File Name     :HDOJ5419.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef unsigned long long int LL;

const int maxn=50500;

/****************BIT***********************/

int n,m;
int w[maxn];
int l[maxn],r[maxn];

inline int lowbit(int x) { return x&(-x); }

int tree[maxn];

void add(int p,int v)
{
	for(int i=p;i<maxn;i+=lowbit(i)) tree[i]+=v;
}

int sum(int p)
{
	int ret=0;
	for(int i=p;i;i-=lowbit(i)) ret+=tree[i];
	return ret;
}
	
LL getC(LL x)
{
	return x*(x-1)/2LL*(x-2)/3LL;
}

LL gcd(LL a,LL b)
{
	if(b==0) return a;
	return gcd(b,a%b);
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++) scanf("%d",w+i);
		memset(tree,0,sizeof(tree));
		for(int i=0;i<m;i++)
		{
			scanf("%d%d",l+i,r+i);
			add(l[i],1); add(r[i]+1,-1);
		}

		if(m<3) { puts("0"); continue; }

		LL up=0,down=getC(m);
		for(int i=1;i<=n;i++)
		{
			LL x=sum(i);
			if(x>=3)
			{
				up=up+w[i]*getC(x);
			}
		}
		if(up==0) { puts("0"); continue; }

		LL g=gcd(up,down);
		if(g==down) cout<<up/g<<endl;
		else cout<<up/g<<"/"<<down/g<<endl;
	}
    
    return 0;
}