文章目录

• 4.26比赛

4.26比赛

NC14734

题目大意：给出12道题做出的概率，问做出0-12道题目的概率

/*

4.26 比赛 ：https://ac.nowcoder.com/acm/problem/14734 

对于每一个题目要么做出，要么做不出，利用dfs进行判断即可 
 // 还有一种dp做法，未完待续......

*/ 

#include<bits/stdc++.h>

using namespace std;

const int N  = 15;

double a[N], b[N], c[N], un[N];
double ans;


void dfs(int pos, int now, int need, double p)  
{
    if(pos == 13) // 当所有题目都写完了 
    {
        if(now == need) { // 达到了需要 
            ans += p;
        }
        return ;
    }  // 判断该题做不做出来 
    if(now < need )
        dfs(pos + 1, now + 1, need, p * (1 - un[pos]));  
    dfs(pos + 1, now, need, p * un[pos]);
}

int main()
{
    for (int i = 1; i <= 12; ++i) cin >> a[i];
    for (int i = 1; i <= 12; ++i) cin >> b[i];
    for (int i = 1; i <= 12; ++i) cin >> c[i];
 	// 一道题做不出来：自己做不出，没听到左边也没听到右边 
    for (int i = 1; i <= 12; ++i) un[i] = (1-a[i]) * (1-b[i]) * (1-c[i]);

    for (int i = 1; i <= 13; ++i)
    {
        ans = 0;
        dfs(1, 0, i - 1, 1);
        printf("%.6f\n", ans);
    }

    return 0;
}

标签：now,int,pos,dfs,ans,need,一题,4.30,4.26
来源： https://blog.csdn.net/ecjtu2020/article/details/116172357