#pragma GCC optimize(3,"Ofast","inline")
#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;
   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]; 
      }
   write(f[n][m]);
	return 0;
}

dp

33分的没有判i<=j的情况

66分的抄洛谷第一篇题解

#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int n, m, x, y, f[N][N];
int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i ++)
		f[i][1] = f[i][0] = 1;
	for (int i = 2; i <= n; i ++)
		for (int j = 2; j <= m; j ++)
			if (i > j) f[i][j] = f[i - 1][j - 1] + f[i - j][j];
            else f[i][j] = f[i - 1][j - 1];
	cout << f[n][m] << '\n';
	return 0;
}