Find the combinatorial number acwing 889. 01 sequence satisfying the condition

Find the combination number AcWing 889. Satisfied 01 Sequence

Original link

AcWing 889. Satisfied 01 Sequence

Algorithm tags

Combinatorial mathematics Combination count Carter LAN number Inverse element Fast power Fermat's small Theorem

Ideas

Code

#include<bits/stdc++.h>
#define int long long
#define abs fabs
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>=b;--i)
using namespace std;
const int N = 100005, mod = 1e9+7;
int pr[N], st[N], s[N], cnt;
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
void put(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar(x%10^48);
}
int qmi(int a, int b, int p){
    int ans=1;
    while(b){
        if(b&1){
            ans=ans*a%p;
        }
        a=a*a%p;
        b>>=1;
    }
    return ans;
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n=read();
    int res=1;
    // Cn 2n 
    Rep(i, 2*n, n+1){
        res=res*i%mod;
    }
    // / n + 1 % p  Demand inverse element 
    rep(i, 1, n+2){
        res=res*qmi(i, mod-2, mod)%mod;
    }
    printf("%lld\n", res);
}

Originality is not easy.
Reprint please indicate the source
If it helps you Don't forget to praise and support