排列组合 概率 DP
这部分还是不太熟悉,网络上的做法
3 3
[1][][] (1,1)=1
[12][][] (2,1)=1/3 [1][2][],[1][][2] (2,2)=2/3
[123][][] (3,1)=1/9 [12][3][],[12][][3] (3,2)=2/9
[13][2][],[1][23][],[13][][2],[1][][23] (3,2)=4/9
[1][2][3],[1][3][2] (3,3)=2/9
//====================================================================||
// ||
// ||
// Author : GCA ||
// 6AE7EE02212D47DAD26C32C0FE829006 ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 55
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
int n,m;
int main() {
ios\_base::sync\_with\_stdio(0);
int ff=0;
while(scanf("%d%d",&n,&m)==2){
double dp\[105\]\[105\];
Set(dp,0);
dp\[0\]\[0\]=1.0;
for(int i=0;i<=n;i++){
for(int j=0;j<=m;j++){
dp\[i+1\]\[j\]+=dp\[i\]\[j\]\*(1.0\*j/m);
dp\[i+1\]\[j+1\]+=dp\[i\]\[j\]\*(1.0\*(m-j)/m);
}
}
printf("Case %d: %.7f\\n",++ff,1-dp\[n\]\[m\]);
}
}