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.
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)
output
0.333333333333 0.333333333333 0.333333333333
output
0.150000000000 0.300000000000 0.550000000000
output
0.057142857143 0.657142857143 0.285714285714
题意：r吃s,s吃p,p吃r,给出3个数数量，求最后3种分别存活的几率
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define eps 1e⑻
typedef __int64 ll;
#define fre(i,a,b) for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define ssf(n) scanf("%s", n)
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define bug pf("Hi
")
using namespace std;
#define INF 0x3f3f3f3f
#define N 105
int r,s,p;
double dp[N][N][N];
double dfs(int pre,int now,int nest) //pre吃now，now吃nest，nest吃pre，最后pre存活的几率
{
if(dp[pre][now][nest]>=0) return dp[pre][now][nest];
if(pre==0) return 0;
if(nest==0) return pre ? 1:0;
double ans=0;
if(pre&&now) ans+=(pre*now*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre,now⑴,nest); //pre吃now
if(now&&nest) ans+=(now*nest*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre,now,nest⑴); //now吃nest，
if(pre&&nest) ans+=(pre*nest*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre⑴,now,nest); //nest吃pre
dp[pre][now][nest]=ans;
return ans;
}
int main()
{
int i,j,n;
mem(dp,⑴);
while(~scanf("%d%d%d",&r,&s,&p))
{
printf("%.10lf
",dfs(r,s,p));
printf("%.10lf
",dfs(s,p,r));
printf("%.10lf
",dfs(p,r,s));
}
return 0;
}