Bag of mice CodeForces -概率DP
阿新 • • 發佈:2018-12-17
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#include<bits/stdc++.h> using namespace std; #define maxn 1111 double dp[2*maxn][maxn]; int w,b; int main() { scanf("%d%d",&w,&b); memset(dp,0.0,sizeof(dp)); for(int i=1; i<=w; i++) { dp[i][0]=1.0; for(int j=1; j<=b; j++) { double x=i; double y=j; //直接獲勝 dp[i][j]+=x/(x+y); //兩個全抓的黑的之後獲勝分兩種情況跑出黑的跑出白的 if(j>=3) dp[i][j]+=y/(x+y)*(y-1.0)/(x+y-1.0)*(y-2.0)/(x+y-2.0)*dp[i][j-3]; if(j>=2) dp[i][j]+=y/(x+y)*(y-1.0)/(x+y-1.0)*(x)/(x+y-2.0)*dp[i-1][j-2]; } } printf("%.9lf\n",dp[w][b]); return 0; }