HDOJ 5645 DZY Loves Balls （概率）

DZY Loves Balls

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 621    Accepted Submission(s): 367

Problem Description

DZY loves playing balls.

He has n balls in a big box. On each ball there is an integer written.

One day he decides to pick two balls from the box. First he randomly picks a ball from the box, and names it A. Next, without putting A back into the box, he randomly picks another ball from the box, and names it B.

If the number written on A is strictly greater than the number on B, he will feel happy.

Now you are given the numbers on each ball. Please calculate the probability that he feels happy.

 

Input

First line contains t denoting the number of testcases.

t testcases follow. In each testcase, first line contains n, second line contains n space-separated positive integers ai, denoting the numbers on the balls.

(1≤t≤300,2≤n≤300,1≤ai≤300)

 

Output

For each testcase, output a real number with 6 decimal places. 

 

Sample Input

231 2 33100 100 100

 

Sample Output

0.5000000.000000
题意：n个球，前取出一个A，不放回取出一个B，求P(A>B)
思路：记录数字出现次数，然后遍历就好了
ac代码：
#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//head int num[333];int main(){int t,i;scanf("%d",&t);while(t--){int n;scanf("%d",&n);mem(num);for(i=0;i<n;i++){int a;scanf("%d",&a);num[a]++;}int k=0;double ans=0.0;for(i=1;i<=300;i++){if(num[i]){double p=(num[i]*1.0)/n;double q=(k*1.0)/((n-1)*1.0);//printf("i= %d p=%lf,q=%lf\n",i,p,q);ans+=p*q;k+=num[i];}}printf("%.6lf\n",ans);}return 0;}