随笔 - 228  文章 - 0 评论 - 7 阅读 - 55413
< 2025年2月 >
26 27 28 29 30 31 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 1
2 3 4 5 6 7 8

E. Fish
time limit per test
3 seconds
memory limit per test
128 megabytes
input
standard input
output
standard output

n fish, numbered from 1 to n, live in a lake. Every day right one pair of fish meet, and the probability of each other pair meeting is the same. If two fish with indexes i and j meet, the first will eat up the second with the probability aij, and the second will eat up the first with the probability aji = 1 - aij. The described process goes on until there are at least two fish in the lake. For each fish find out the probability that it will survive to be the last in the lake.

Input

The first line contains integer n (1 ≤ n ≤ 18) — the amount of fish in the lake. Then there follow n lines with n real numbers each — matrix aaij (0 ≤ aij ≤ 1) — the probability that fish with index i eats up fish with index j. It's guaranteed that the main diagonal contains zeros only, and for other elements the following is true: aij = 1 - aji. All real numbers are given with not more than 6 characters after the decimal point.

Output

Output n space-separated real numbers accurate to not less than 6 decimal places. Number with index i should be equal to the probability that fish with index i will survive to be the last in the lake.

Sample test(s)
input
2
0 0.5
0.5 0
output
0.500000 0.500000 
input
5
0 1 1 1 1
0 0 0.5 0.5 0.5
0 0.5 0 0.5 0.5
0 0.5 0.5 0 0.5
0 0.5 0.5 0.5 0
output
1.000000 0.000000 0.000000 0.000000 0.000000 

复制代码
 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 using namespace std;
 5 double a[18][18],b[1<<18];
 6 int fun(int x)
 7 {
 8     int s=0;
 9     while(x)
10     {
11         s+=x&1;
12         x>>=1;
13     }
14     return s;
15 }
16 int main()
17 {
18     int n,i,r,t,j;
19     cin>>n;
20     for(i=0; i<n; i++)for(j=0; j<n; j++)scanf("%lf",&a[i][j]);
21     memset(b,0,sizeof(b));
22     b[(1<<n)-1]=1;
23     for(i=(1<<n)-1; i>=0; i--)
24     {
25         int c=fun(i);
26         c=c*(c-1)/2;
27         for(r=0;r<n;r++)
28         if(i&(1<<r))
29         for(t=0;t<n;t++)
30         if(i&(1<<t))
31         b[i-(1<<t)]+=b[i]*a[r][t]/c;
32     }
33     for(r=0;r<n-1;r++)
34     printf("%.6lf ",b[1<<r]);printf("%.6lf\n",b[1<<r]);
35 }
View Code
复制代码

 

posted on   ERKE  阅读(239)  评论(0编辑  收藏  举报
编辑推荐:
· .NET Core内存结构体系(Windows环境)底层原理浅谈
· C# 深度学习:对抗生成网络(GAN)训练头像生成模型
· .NET 适配 HarmonyOS 进展
· .NET 进程 stackoverflow异常后,还可以接收 TCP 连接请求吗?
· SQL Server统计信息更新会被阻塞或引起会话阻塞吗?
阅读排行:
· 本地部署 DeepSeek:小白也能轻松搞定!
· 传国玉玺易主,ai.com竟然跳转到国产AI
· 自己如何在本地电脑从零搭建DeepSeek!手把手教学,快来看看! (建议收藏)
· 我们是如何解决abp身上的几个痛点
· 如何基于DeepSeek开展AI项目
点击右上角即可分享
微信分享提示