I
don’t know how to count. Probability and Mathematical Statistics in the next semester.
E.Shorten ABC
When the B array is determined, it is not difficult to find that the question asked is the number of plans for the A sequence selected from the B sequence.
It is equivalent to selecting s+n balls from m+n. We assume that the n balls of multiple selections are a partition, and the array A is separated. The partition of the multiple selection determines the ball in B The number (because most m) the
answer is C m + ns + n, s C_{m+n}^{s+n},sCm+ns+n,s is the total number of balls in array A.
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=200010;
const ll mod=1e9+7;
ll qmi(ll a,ll b,ll p)
{
ll res=1;
while(b)
{
if(b&1) res=res*a%p;
b>>=1;
a=a*a%p;
}
return res;
}
ll C(ll n,ll m)
{
ll res=1;
for(int i=1;i<=m;i++)
res=res*qmi(i,mod-2,mod)%mod*(n-i+1)%mod;
return res;
}
int main()
{
IO;
int T=1;
//cin>>T;
while(T--)
{
int n,m;
cin>>n>>m;
ll s=0;
for(int i=1;i<=n;i++)
{
int a;
cin>>a;
s+=a;
}
cout<<C(m+n,s+n)<<'\n';
}
return 0;
}
Dig a hole first