题意:n<20个箱子,每个里面有fi朵颜色相同的花,不同箱子里的花颜色不同,要求取出s朵花,问方案数

题解:假设不考虑箱子的数量限制,隔板法可得方案数是c(s+n-1,n-1),当某个箱子里的数量超过fi时,方案数是c(s-f[i]-1+n-1,n-1),容斥原理求,状压枚举哪几个箱子超过了f[i],答案就是超过0个-超过1个+超过2个...

由于c(n,m)的m很小,直接暴力求解

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

//#include <bits/extc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define mt make_tuple

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 1000000007

#define ld long double

//#define C 0.5772156649

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

#define ull unsigned long long

#define bpc __builtin_popcount

//#define base 1000000000000000000

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

//using namespace __gnu_pbds;

const ull ba=233;

const db eps=1e-5;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=100000+10,maxn=1000000+10,inf=0x3f3f3f3f;

ll s,a[30],inv[30];

ll c(ll a,ll b)

{

    ll ans=1;

    for(ll i=a;i>=a-b+1;i--)ans=ans*(i%mod)%mod;

    return ans*inv[b]%mod;

}

int main()

{

    inv[0]=1;

    for(int i=1;i<30;i++)inv[i]=inv[i-1]*qp(i,mod-2)%mod;

    int n;

    scanf("%d%lld",&n,&s);

    for(int i=0;i<n;i++)scanf("%lld",&a[i]);

    ll ans=0;

    for(int i=0;i<(1<<n);i++)

    {

        int f=0;

        ll sum=0;

        for(int j=0;j<n;j++)if((i>>j)&1)

        {

            sum+=a[j]+1;

            f^=1;

        }

        if(sum>s)continue;

        if(!f)add(ans,c(s-sum+n-1,n-1));

        else sub(ans,c(s-sum+n-1,n-1));

    }

    printf("%lld\n",ans);

    return 0;

}

/********************

********************/