HDU 5419 Victor and Toys （）

Victor and Toys

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 467    Accepted Submission(s): 157

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

13 41 1 52 31 33 31 1

 

Sample Output

5/4

 

Source

BestCoder Round #52 (div.2)

 

问题描述

Victor有$n$个玩具，编号为$1$到$n$，其中编号为$i$的玩具有w_iw​i​​点有趣值。Victor对数字特别敏感，他在头脑中生成了$m$个区间，其中第$i$个区间为[l_i,r_i][l​i​​,r​i​​]，并随机选取了$3$个互不相同的数$i,j,k(1\le i<j<k\le m)$，将所有满足\max(l_i,l_j,l_k)\leq x\leq \min(r_i,r_j,r_k)max(l​i​​,l​j​​,l​k​​)≤x≤min(r​i​​,r​j​​,r​k​​)的编号为$x$的玩具取出，他想知道他取出的玩具的有趣值之和的期望是多少，你能帮帮他吗？

输入描述

第一行包含一个整数$T$，表示测试数据的组数。每组测试数据的第一行有两个整数$n$, $m$，表示玩具的个数和区间的个数。接下来一行有$n$个数，其中第$i$个数为w_iw​i​​。接下来$m$行，每行两个整数，其中第$i$行表示l_il​i​​, r_ir​i​​。$1\le T\le 10$。$1\le n,m\le 50000$。1\leq w_i\leq 51≤w​i​​≤5。1\leq l_i\leq r_i\leq n1≤l​i​​≤r​i​​≤n。

输出描述

每组测试数据输出一行，若答案为整数，则输出一个整数，否则输出形如$p/q$的最简分数。

输入样例

13 41 1 52 31 33 31 1

输出样例

5/4

解题思路：和的期望 = (∑任意一种挑选方式的和x)/(满足i<=j<=k的挑选方式的总数y) = x / y 。分母y 就是m个数中取3个数的组合C(m,3)，即y = m*(m-1)*(m-2)/6 ，对于分子x，如果第i个数出现的区间数是num（简言之就是有num个区间包含第i个数），那么第i个数对分子的贡献值是w[i]*C(num,3)。因此枚举i，累加和。（预处理过程与HDU 5124 line 类似）

代码如下：

#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <cmath>#include <string>#include <algorithm>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#include <ctime>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-6)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1000000007using namespace std;typedef long long ll;typedef unsigned long long ULL;int w[50005],dp[50005];ll gcd(ll a,ll b){    return b==0?a:gcd(b,a%b);}int main(){    ll i,j,k,n,m,l,r,t;    scanf("%I64d",&t);    while(t--)    {        scanf("%I64d%I64d",&n,&m);        for(i=1;i<=n;i++)            scanf("%I64d",&w[i]);        memset(dp,0,sizeof(dp));        for(i=0;i<m;i++)        {            scanf("%I64d%I64d",&l,&r);            dp[l]++;            dp[r+1]--;        }        ll x=0;        ll num=0;        for(i=1;i<=n;i++)        {            num+=dp[i];            if(num>=3)                x+=num*(num-1)*(num-2)/6*w[i];        }        ll y=m*(m-1)*(m-2)/6;        //printf("x=%I64d y=%I64d\n",x,y);        ll d=gcd(x,y);        if(d)        {            x=x/d;            y=y/d;            if(y==1)                printf("%I64d\n",x);            else                printf("%I64d/%I64d\n",x,y);        }        else            printf("0\n");    }    return 0;}