#include<bits/stdc++.h>
using namespace std;
int a[305][305],n,k;
int dfs(int x,int y){
	if(y>x)return 0;
	if(x==y)return 1;
	if(y==1)return 1;
	if(a[x][y]!=-1)return a[x][y];
	return a[x][y]=dfs(x-y,y)+dfs(x-1,y-1);
}
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	memset(a,-1,sizeof(a));
	scanf("%d%d",&n,&k);
	printf("%d",dfs(n,k));
	return 0;
}

简单dp
#include<bits/stdc++.h>
using namespace std;
long f[1005][1005],i,j,n,m;
long read()
{
	long x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')
	{
		x=x*10+(ch-'0');
		ch=getchar();
	}
	return x*f;
}
void write(long x) {
	if(x<0){
		putchar('-');
		write(-x);
		return;
	}
	if(x>=10)write(x/10);
	putchar(x%10+'0');
}
int main()
{
   n=read(),m=read();
   for(i=1;i<=n;i++)f[i][0]=f[i][1]=1;//初始值
   for(i=2;i<=n;i++)
      for(j=2;j<=m;j++)
      {
         f[i][j]=f[i-1][j-1];
         if(i>j)f[i][j]+=f[i-j][j]; //简单的动态转移方程
      }
   write(f[n][m]);
	return 0;
}