AcWing 885. 求组合数 I · znzryb

## 思路讲解

![image](https://media.znzryb.com/20250330-788c5003/image_ca3732c5.png)

这个递推法其实就是把两种情况分开来考虑了（即选一个人的情况+不选一个人的情况）

## AC代码

[https://www.acwing.com/problem/content/submission/code_detail/40911169/](https://www.acwing.com/problem/content/submission/code_detail/40911169/)

```cpp
// Problem: 求组合数 I
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/description/887/
// Memory Limit: 64 MB
// Time Limit: 1000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(2e3)+15,INF=static_cast<ll>(5e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;

ll N,M,T,C[MAXN][MAXN];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	N=2010;
	C[1][1]=1;
	C[0][1]=1;
	FOR(i,1,N){
		C[0][i]=1;
	}
	FOR(i,2,N){
		FOR(j,1,i){
			C[j][i]=(C[j-1][i-1]+C[j][i-1])%mod;
		}
	}
	cin>>T;
	FOR(i,1,T){
		ll a,b;
		cin>>a>>b;
		cout<<C[b][a]<<"\n";
	}
	return 0;
}
/*
AC
https://www.acwing.com/problem/content/submission/code_detail/40911169/
*/
```

## 心路历程（WA，TLE，MLE……）