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CF# 301 D Bad Luck Island(概率dp+记忆化)

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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 rs 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

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; }




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