dp+前缀和，dp[i][j]代表使用 j 个正整数组成 i 的方案数，最终输出dp[2021][5]即可。

#include <bits/stdc++.h>
using namespace std;

#define ll long long

ll dp[2022][6];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);

	for (int i = 1; i <= 2021; ++ i) {
		dp[i][1] = 1;
	}

  for (int j = 2; j <= 5; j++) {
      for (int i = 1; i <= 2021; i++) {
          long long sum = 0;
          for (int x = 1; x <= i - j + 1; x++) {
              sum += dp[i - x][j - 1];
          }
          dp[i][j] = sum;
      }
  }
  
  cout << dp[2021][5];

	return 0;
}