Codeforces Round #258 (Div. 2) 容斥+Lucas

题目链接：

http://codeforces.com/problemset/problem/451/E

E. Devu and Flowers

time limit per test4 secondsmemory limit per test256 megabytes
#### 问题描述
> Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.
>
> Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109 + 7).
>
> Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.

输入

The first line of input contains two space-separated integers n and s (1 ≤ n ≤ 20, 0 ≤ s ≤ 1014).

The second line contains n space-separated integers f1, f2, ... fn (0 ≤ fi ≤ 1012).

输出

Output a single integer — the number of ways in which Devu can select the flowers modulo (109 + 7).

样例输入

3 5

1 3 2

样例输出

3

题意

给你n个花瓶，每个花瓶上装有f[i]朵颜色相同的玫瑰。现在你要从中取s朵，问有多少种取法。

题解

首先，如果每个花瓶里面的花都有无限朵的话，那么这就是简单的隔板问题(c[s+n-1][n-1])，那么我们可以通过容斥把问题都转换成无限制的问题。 首先我们考虑第i个超过了限制，那么相当于对剩下的sum=s-f[i]-1，做一遍无限制的隔板，这样我们对n个花瓶容斥一遍，就能求出答案。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int mod=1e9+7;

LL n;
LL s,f[22];

void gcd(LL a,LL b,LL& d,LL& x,LL& y){
    if(!b){ d=a; x=1; y=0; }
    else{
        gcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}

LL Inv(LL a,LL n){
    LL d,x,y;
    gcd(a,n,d,x,y);
    return d==1?(x+n)%n:-1;
}

LL get_C(LL n,LL m){
    LL a=1,b=1;
    for(int i=1;i<=m;i++){
        b=b*i%mod;
        a=a*(n-i+1)%mod;
    }
    return a*Inv(b,mod)%mod;
}

LL Lucas(LL n,LL m){
    if(m==0) return 1;
    return get_C(n%mod,m%mod)*Lucas(n/mod,m/mod)%mod;
}

int main() {
    scf("%I64d%I64d",&n,&s);
    for(int i=0;i<n;i++) scf("%I64d",f+i);
    LL ans=0;
    for(int i=0;i<(1<<n);i++){
        LL sum=s;
        int cnt=0;
        for(int j=0;j<n;j++){
            if((i&(1<<j))==0) continue;
            cnt++;
            sum-=(f[j]+1);
        }
        if(sum>=0){
            if(cnt&1){
                ans-=Lucas(sum+n-1,n-1);
            }else{
                ans+=Lucas(sum+n-1,n-1);
            }
            ans=(ans%mod+mod)%mod;
        }
    }
    prf("%I64d\n",ans);
    return 0;
}

/*
5 3
1 2 9
3 4 5
3 5 3
*/