首次自解 DP，因为 64 位元 WA 了一次
//============================================================================
// Name : Bar Codes.cpp
// Date : 2013/4/12 上午12:15:02
// Author : GCA
//============================================================================
#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 MM 155
#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);
}
inline Int Imax(Int x,Int y){
if(x>y)return x;
else return y;
}
Int dp\[MM\]\[MM\];
Int N,K,M;
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%lld%lld%lld",&N,&K,&M)){
Set(dp,0);
for(int i=1;i<=M;i++){
dp\[1\]\[i\]=1;
}
for(Int i=2;i<=K;i++){
for(Int j=i;j<=N;j++){
Int sum=0;
for(Int k=j-1;k>=Imax(j-M,1);k--){
sum+=dp\[i-1\]\[k\];
}
dp\[i\]\[j\]=sum;
}
}
printf("%lld\\n",dp\[K\]\[N\]);
}
}