传送门
解题思路:第i组的人数必须大于Ci，于是我们可以将问题转化为\(N-\sum_{i=1}^M Ci\)人分到M组中，且保证每一组的人数大于0，然后我们可以使用隔板法求出分的的组数\(C_{N-1-\sum_{i=1}^M Ci}^{m-1}\)
我们可以直接通过基本的组合公式+费马小定理直接求，也可以通过Lucas定理求得:
直接求:
Code:

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
ll n,m,a,t;ll power(ll a,ll b) {ll ans = 1;while(b) {if(b&1) ans = (ans*a)%mod;a = (a*a)%mod;b>>=1;}return ans;
}ll inv(ll x) {return power(x,mod-2);
}ll C(ll a,ll b) {ll ans =  1;for(ll i = 1;i <= b; ++i) {ans = (ans*(a-i+1)) % mod;ans = (ans*inv(i)) % mod;}return ans % mod;
}int main()
{while(~scanf("%lld%lld",&n,&m)) {a = 0LL;for(int i = 0;i < m; ++i) {scanf("%lld",&t);a += t;}printf("%lld\n",C(n-a-1,m-1));}return 0;
}

Lucas定理:

#include <iostream>
#include <cstdio>
using namespace std;#define MOD 1000000007
typedef long long LL;int Read() { //快读char c; int ans(0);while(!isdigit(c = getchar()));do ans = ans * 10 + c - 48;while(isdigit(c = getchar()));return ans;
}int n, m;LL Qpow(LL a, LL b, LL p) { //快速幂 LL ans = 1;for(; b; b>>=1, (a*=a) %= p)if(b & 1) (ans *= a) %= p;return ans;
}LL c(LL n, LL m, LL p) {if(n < m) return 0;if(m > n - m) m = n - m;LL s1 = 1, s2 = 1;for(int i=0; i<m; i++) {s1 = s1 * (n - i) % p;s2 = s2 * (i + 1) % p;}return s1 * Qpow(s2, p-2, p) % p;
}LL Lucas(LL n, LL m, LL p) { //Lucasif(m == 0) return 1;return c(n % p, m % p, p) * Lucas(n / p, m / p, p) % p;
}int main() {n = Read(), m = Read();for(int i=1; i<=m; i++) n -= Read();printf("%d\n", Lucas(n-1, m-1, MOD));return 0;
}