Random Events C++

Random Events
比赛题目  题目统计 全部提交 时间限制：C/C++ 2000MS，其他语言 4000MS
内存限制：C/C++ 256MB，其他语言 512MB

描述

 

Ron is a happy owner of a permutation a of length n.

A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Ron's permutation is subjected to m experiments of the following type: (ri​, pi​). This means that elements in range [1,ri​] (in other words, the prefix of length ri​) have to be sorted in ascending order with the probability of pi​. All experiments are performed in the same order in which they are specified in the input data.

As an example, let's take a look at a permutation [4,2,1,5,3] and an experiment (3,0.6). After such an experiment with the probability of 60% the permutation will assume the form [1,2,4,5,3] and with a 40% probability it will remain unchanged.

You have to determine the probability of the permutation becoming completely sorted in ascending order after m experiments.

 

输入描述

 

Each test contains one or more test cases. The first line contains the number of test cases t (1≤t≤100).

The first line of each test case contains two integers n and m (1≤n,m≤105) — the length of the permutation and the number of experiments, respectively.

The second line of each test case contains n integers a1​,a2​,…,an​ (1≤ai​≤n) — contents of the permutation.

The following m lines of each test case each contain an integer ri​ and a real number pi​ (1≤ri​≤n,0≤pi​≤1) — the length of the prefix and the probability of it being sorted. All probabilities are given with at most 6 decimal places.

It is guaranteed that the sum of n and the sum of m does not exceed 105 (∑n,∑m≤105).

 

输出描述

 

For each test case, print a single number — the probability that after all experiments the permutation becomes sorted in ascending order. Your answer will be considered correct if its absolute or relative error does not exceed 10−6.

 

用例输入 1 

4
4 3
4 3 2 1
1 0.3
3 1
4 0.6
5 3
4 2 1 3 5
3 0.8
4 0.6
5 0.3
6 5
1 3 2 4 5 6
4 0.9
5 0.3
2 0.4
6 0.7
3 0.5
4 2
1 2 3 4
2 0.5
4 0.1

用例输出 1 

0.600000
0.720000
0.989500
1.000000

提示

 

Explanation of the first test case: It can be demonstrated that whether the final permutation is sorted or not depends solely on sorting being performed in the (4,0.6) experiment.

 

 

题目：这是一个充满恶意的东西：

要找一个升序数组的概率

大致意思就是先输入t，循环t次，然后输入n和m代表接下来的数组大小以及案例个数，然后随后就是ai（i=0,1,2,3,...n-1)，换行后输入案例分为index（下标）和rate（概率）意思是在这个index之前的数列有rate的几率变为升序排列，也有1-rate没法变为升序，这里就需要我们先排序知道我们最少需要改变的位置是哪里，就比如输入数列为1，4，3，2，5；排序后就是1，2，3，4，5我们知道我们需要排序在4位置之前，我们只用处理index>=4的情况。

这时候给出了

3 0.6

4 0.4

5 0.1

能够变为升序的可能性就是1-（（1-0.4）*（1-0.1））=0.46或者0.4+（1-0.4）*0.1=0.46(这一种写法有点复杂点，就没写)

所以我们可以知道这个题目很简单的概率就这样写：

AC：

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;


int main(){
    int t;
    cin>>t;
    while(t--){
        int n,m;
        cin>>n>>m;
        int aa[n],bb[n];
        for(int i=0;i<n;i++){
            cin>>aa[i];
            bb[i]=aa[i];
        }
        sort(aa,aa+n);//排序是为了知道我们需要找到的下标为多少
        int mark=0;
　　　　　//int mark=n-1;
　　　　　//while(aa[mark]==bb[mark]&&mark>=1)mark--;//是不是很简洁这个while，很灵性，比for好看多了

        for(int i=n-1;i>=0;i--)
            if(aa[i]!=bb[i]){
                mark=i+1;
                break;
            }
        double ans=1;
        for(int i=0;i<m;i++){
            int index;
            double rate;
            cin>>index>>rate;
            if(index>=mark) ans*=(1-rate);
        }
        ans=1.0-ans;
        if(mark) printf("%.6f\n",ans);
        else cout<<"1.000000"<<endl;
    }
    return 0;
}

 

标签：aa,probability,int,Random,C++,mark,permutation,test,Events
来源： https://www.cnblogs.com/jerrytangcaicai/p/16279083.html