D. Bad Luck Island
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
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.
Output
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.
Sample test(s)
input
2 2 2
output
0.333333333333 0.333333333333 0.333333333333
input
2 1 2
output
0.150000000000 0.300000000000 0.550000000000
input
1 1 3
output
0.057142857143 0.657142857143 0.285714285714
链接：http://codeforces.com/contest/540/problem/D
题意：一个岛上有石头人，剪刀人，布人，每天会有两个人相遇，根据相克会死掉一个人。问最后只剩下石头人的几率，只剩剪刀人的几率，布人的几率。
做法：dp[i][j][k] 代表有i个石头，j个剪刀，k个布的几率。
以剪刀和布相遇为例，会有转移 dp[i][j][k-1]+=dp[i][j][k]*j*k/(i+j+k)/(i+j+k-1) 。
但是这是不够的，因为还有平局的情况。平局的时候，状态又转移回了dp[i][j][k]，又从原状态开始转移，所以转移的比例还是一样的。
所以可以直接把 所有的转移几率相加，然后在转移的时候除掉。转移方程变为dp[i][j][k-1]+=dp[i][j][k]*j*k/(i+j+k)/(i+j+k-1)/tem;
tem为不是平局的几率总和。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64d
#define pi acos(-1.0)
double dp[110][110][110];
int main()
{
int shi,jian,bu;
while(scanf("%d%d%d",&shi,&jian,&bu)!=EOF)
{
memset(dp,0,sizeof dp);
dp[shi][jian][bu]=1;
for(int i=shi;i>=0;i--)
{
for(int j=jian;j>=0;j--)
{
for(int k=bu;k>=0;k--)
{
//printf("%d %d %d %lf\n ",i,j,k,dp[i][j][k]);
if(i+j==0||i+k==0||j+k==0)
continue;
double tem=0;
if(k!=0)
tem+=1.0*j*k/(i+j+k)/(i+j+k-1);
if(j!=0)
tem+=1.0*j*i/(i+j+k)/(i+j+k-1);
if(i!=0)
tem+=1.0*i*k/(i+j+k)/(i+j+k-1);
if(k!=0)
dp[i][j][k-1]+=dp[i][j][k]*j*k/(i+j+k)/(i+j+k-1)/tem;
if(j!=0)
dp[i][j-1][k]+=dp[i][j][k]*j*i/(i+j+k)/(i+j+k-1)/tem;
if(i!=0)
dp[i-1][j][k]+=dp[i][j][k]*i*k/(i+j+k)/(i+j+k-1)/tem;
}
}
}
double aj=0;
double as=0,ab=0;
for(int i=1;i<=jian;i++)
{
aj+=dp[0][i][0];
}
for(int i=1;i<=shi;i++)
{
as+=dp[i][0][0];
}
for(int i=1;i<=bu;i++)
{
ab+=dp[0][0][i];
}
printf("%.9lf %.9lf %.9lf\n",as,aj,ab);
}
return 0;
}