D. Bag of mice

原题链接

题解

注意数据范围，设 $dp[x][y]$ 为x只白鼠，y只黑鼠，A先手赢的概率

则每次选取要么A赢，要么B赢，要么都没有赢进入下一轮

code

#include<bits/stdc++.h>
using namespace std;
/*
#define int long long
#define double long double
#define lowbit(x) ((x)&(-x))
const int inf=1e18;
const int mod=1e9+7;

const int N=4e5;
int qpow(int a,int n)
{
    int res=1;
    while(n)
    {
        if(n&1) res=res*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return res;
}
int inv(int x)
{
    return qpow(x,mod-2);
}
int fa[2000005];
int finds(int now){return now==fa[now]?now:finds(fa[now]);}

vector<int> G[200005];

int dfn[200005],low[200005];
int cnt=0,num=0;
int in_st[200005]={0};
stack<int> st;
int belong[200005]={0};

void scc(int now,int fa)
{
    dfn[now]=++cnt;
    low[now]=dfn[now];
    in_st[now]=1;
    st.push(now);

    for(auto next:G[now])
    {
        if(next==fa) continue;

        if(!dfn[next])
        {
            scc(next,now);
            low[now]=min(low[now],low[next]);
        }
        else if(in_st[next])
        {
            low[now]=min(low[now],dfn[next]);
        }
    }

    if(low[now]==dfn[now])
    {
        int x;
        num++;
        do
        {
            x=st.top();
            st.pop();
            in_st[x]=0;
            belong[x]=num;
        }while(x!=now);
    }
}
vector<int> prime;
bool mark[200005]={0};
void shai()
{
    for(int i=2;i<=200000;i++)
    {
        if(!mark[i]) prime.push_back(i);

        for(auto it:prime)
        {
            if(it*i>200000) break;

            mark[it*i]=1;
            if(it%i==0) break;
        }
    }
}
*/
double dp[1005][1005];
void solve()
{
    int a,b;
    cin>>a>>b;
    memset(dp,0,sizeof dp);
    for(int x=1;x<=a;x++) dp[x][0]=1;

    for(int x=1;x<=a;x++)
    {
        double xx=x;
        dp[x][1]=xx/(xx+1);
        dp[x][2]=xx/(xx+2)+2.0/(xx+2.0)*1.0/(xx+1.0)*dp[x-1][0];
        for(int y=3;y<=b;y++)
        {
            double yy=y;
            dp[x][y]=xx/(xx+yy)+yy/(xx+yy)*(yy-1)/(xx+yy-1)*((yy-2)/(xx+yy-2)*dp[x][y-3]+xx/(xx+yy-2)*dp[x-1][y-2]);
        }
    }

    printf("%.9lf",dp[a][b]);
}
signed main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int TT=1;
    //cin>>TT;
    while(TT--) solve();
    return 0;
}