Recursion-the number of integer partitions

Title description:
Algorithm idea:

• The maximum number of zeros of n can be set as the number of division formulas that do not exceed m as q(n,m)
• Initial conditions: q(n,0)=0,q(n,1)=1,q(1,n)=1

	for(int i=1;i<=s;i++){
    
    
		q[i][0]=0;
		q[i][1]=1;
		q[1][i]=1;
	}

• Get the recurrence relationship: q(n,m)=q(n,m-1)+q(nm,m) when nm>m, q(n,m)=q(n,m- when nm<=m 1)+q(nm,nm)

	for(int i=2;i<=s;i++){
    
    
		for(int j=1;j<=i-1;j++){
    
    
			// 零数大于自身数 
			if(i-j<j) q[i-j][j] = q[i-j][i-j];
			// 最大零数小于m的划分式个数+等于m的划分式个数 
			q[i][j] = q[i][j-1] + q[i-j][j];     
		}
		// 自身也是划分数 如q(5,5) = 1+q(5,4),这个1表示5=5
		q[i][i] = q[i][i-1] + 1; 
	}

All codes:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 1e3;
long long q[N][N];
int n,m,s;

int main(){
    
    
	cin>>s;
	// 初始化 
	for(int i=1;i<=s;i++){
    
    
		q[i][0]=0;
		q[i][1]=1;
		q[1][i]=1;
	}
	for(int i=2;i<=s;i++){
    
    
		for(int j=1;j<=i-1;j++){
    
    
			// 零数大于自身数 
			if(i-j<j) q[i-j][j] = q[i-j][i-j];
			// 最大零数小于m的划分式个数+等于m的划分式个数 
			q[i][j] = q[i][j-1] + q[i-j][j];     
		}
		// 自身也是划分数 如q(5,5) = 1+q(5,4),这个1表示5=5
		q[i][i] = q[i][i-1] + 1; 
	}
	cout<<q[s][s]<<endl; 
}