浙工大姗姗杯round2 G
来源:互联网 发布:安居客网络经纪人登录 编辑:程序博客网 时间:2024/04/28 05:38
D. Bad Luck Island
The Bad Luck Island is inhabited by three kinds of species: r rocks, s scissors and p papers. At some moments of time two random individuals meet (all pairs of individuals can meet equiprobably), and if they belong to different species, then one individual kills the other one: a rock kills scissors, scissors kill paper, and paper kills a rock. Your task is to determine for each species what is the probability that this species will be the only one to inhabit this island after a long enough period of time.
The single line contains three integers r, s and p (1 ≤ r, s, p ≤ 100) — the original number of individuals in the species of rock, scissors and paper, respectively.
Print three space-separated real numbers: the probabilities, at which the rocks, the scissors and the paper will be the only surviving species, respectively. The answer will be considered correct if the relative or absolute error of each number doesn't exceed 10 - 9.
2 2 2
0.333333333333 0.333333333333 0.333333333333
2 1 2
0.150000000000 0.300000000000 0.550000000000
1 1 3
0.057142857143 0.657142857143 0.285714285714
思路:概率dp。本来以为有公式可以推到,但是想想不可能。dp[i][j][k],i,j,k分别表示包剪锤数量,i*j+i*k+j*k是总的相遇情况,又因为两种物种相遇死亡方已知,所以可以用i*j/(i*j+i*k+j*k)来表示概率。接下来就是dp
#include <stdio.h>#include <iostream>using namespace std;double dp[128][128][128];int main(){int r, s, p;cin >> r >> s >> p;for(int i = 1; i <= 127; i++) for(int j = 0; j <= 127; j++) dp[i][j][0] = 1;for(int i = 1; i <= 127; i++) for(int j = 1; j <= 127; j++) for(int k = 1; k <= 127; k++){ double p = i*j + j*k + i*k;//总情况数; dp[i][j][k] += i*k*dp[i - 1][j][k]; dp[i][j][k] += i*j*dp[i][j - 1][k]; dp[i][j][k] += j*k*dp[i][j][k - 1]; dp[i][j][k] /= p; } printf("%.12lf %.12lf %.12lf\n", dp[r][s][p], dp[s][p][r], dp[p][r][s]);return 0;}
- 浙工大姗姗杯round2 G
- 浙工大姗姗杯round2 CodeForces 103BCthulhu
- 浙工大姗姗杯round2 CodeForces 116BLittle Pigs and Wolves
- 浙工大姗姗杯round1 G
- 浙工大姗姗杯round1 A
- 浙工大姗姗杯round1 C
- 浙工大姗姗杯round1 D
- 浙工大姗姗杯round1 E
- 浙工大姗姗杯round1 F
- 浙工大姗姗杯round3 A
- 新生训练赛round2--G. Maya Calendar(模拟)
- Round2 - Probability
- [弱校胡策Round2]
- SDOI2014 Round2 Day1
- SDOI2014 Round2 Day2
- SDOI2015 Round2 Day1总结
- NOIP四校联训Round2小结
- 四校联训Round2心得体会
- 还没太搞明白怎么用
- swing GeneralPath::cubicTo绘制平滑曲线
- 例外的种类
- linux安装jdk
- Android Hook框架adbi的分析(3)---编译和inline Hook实践
- 浙工大姗姗杯round2 G
- iOS MD5 和 Base64 算法
- 判断点是否在线段上
- Annotation
- nginx的url重写rewrite模块
- css解决input和select的宽度不一致问题
- JAVA线程池应用实践
- 母函数入门
- 盒子模型