bzoj 1419: Red is good

f[i][j]表示有i个R,j个B的期望，然后转移。

注意输出多位小数下取整，小数点之后要06lld

ll ans=floor(f[now][B]*1000000);

printf("%lld.%06lld\n",ans/1000000,ans%1000000);

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
#define inf 1e9
#define eps 1e-8
#define md
using namespace std;
double f[2][5010];
int main()
{
int R,B;
scanf("%d%d",&R,&B);
int pre=0,now=1;
for (int i=1;i<=R;i++)
{
swap(now,pre);
f[now][0]=i;
for (int j=1;j<=B;j++)
{
double p=1.0*i/(i+j),g=(f[pre][j]+1)*p+(f[now][j-1]-1)*(1-p);
if (g<0) f[now][j]=0; else f[now][j]=g;
}
}
ll ans=floor(f[now][B]*1000000);
printf("%lld.%06lld\n",ans/1000000,ans%1000000);
return 0;
}