首页 > 代码库 > CodeForces 148D Bag of mice

CodeForces 148D Bag of mice

概率,$dp$。

$dp[i][j][0]$表示还剩下$i$个白猫,$j$个黑猫,公主出手的情况下到达目标状态的概率。

$dp[i][j][1]$表示还剩下$i$个白猫,$j$个黑猫,龙出手的情况下到达目标状态的概率。

一开始$dp[i][0][0]$均为$1$,答案为$dp[w][b][0]$。递推式很容易写:

$dp[i][j][0]=i/(i+j)+j/(i+j)*dp[i][j-1][1]$

$dp[i][j][1]=j/(i+j)*i/(i+j-1)*dp[i-1][j-1][0]+j*(j-1)/((i+j)*(i+j-1))*dp[i][j-2][0]$

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){    freopen("D:\\in.txt","r",stdin);    freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){    char c = getchar();    x = 0;    while(!isdigit(c)) c = getchar();    while(isdigit(c))    {        x = x * 10 + c - 0;        c = getchar();    }}double dp[1005][1005][2];int w,b;int main(){    while(~scanf("%d%d",&w,&b))    {        memset(dp,0,sizeof dp);        for(int i=1;i<=w;i++) dp[i][0][0]=1;        for(int i=1;i<=w;i++)        {            for(int j=1;j<=b;j++)            {                dp[i][j][0]=1.0*i/(i+j)+1.0*j/(i+j)*dp[i][j-1][1];                dp[i][j][1]=1.0*j/(i+j)*i/(i+j-1)*dp[i-1][j-1][0];                if(j>=2) dp[i][j][1]+=1.0*j*(j-1)/((i+j)*(i+j-1))*dp[i][j-2][0];            }        }        printf("%.9f\n",dp[w][b][0]);    }    return 0;}

 

CodeForces 148D Bag of mice