AcWing 885. Find the combination number I (recursive preprocessing)

Problem plane connection

https://www.acwing.com/problem/content/887/

Ideas

Through the knowledge of combinatorial mathematics, we can know $C_a^b=C_{a-1}^b+C_{a-1}^{b-1}$ Of , The right side of the equation is from $a-1$ Choose from the number of $b-1$ Number and from $a-1$ Choose from the number of $b$ Number of cases , Corresponding to the current $b$ Number of options and no options （ It seems to be a little similar to the idea of backpack ）, Then we can get the result by recursion 2000 All combinations within , See code for details

Code

#include<bits/stdc++.h>
using namespace std;
//---------------- Custom part ----------------
#define ll long long
#define mod 1000000007
#define endl "\n"
#define PII pair<int,int>
#define INF 0x3f3f3f3f

int dx[4] = {
    -1, 0, 1, 0}, dy[4] = {
    0, 1, 0, -1};

ll ksm(ll a,ll b) {
    
	ll ans = 1;
	for(;b;b>>=1LL) {
    
		if(b & 1) ans = ans * a % mod;
		a = a * a % mod;
	}
	return ans;
}

ll lowbit(ll x){
    return -x & x;}

const int N = 2e3+10;
//---------------- Custom part ----------------
ll t,n,m,q,c[N][N];

void init(){
    
	for(int i = 0;i < N; ++i) c[i][0] = 1;// Initialization selection 0 The situation of 
	for(int i = 1;i < N; ++i) {
    
		for(int j = 1;j <= i; ++j) {
    
			c[i][j] = (c[i-1][j] + c[i-1][j-1]) % mod;
		}
	}
}
void slove(){
    
	ll a,b;
	cin>>a>>b;
	cout<<c[a][b]<<endl;
}

int main()
{
    
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	init();
	cin>>t;
	while(t--){
    
		slove();
	}
	
	return 0;
}