标签：line any sig ast int which lin UI blog
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.
Input
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).
Output
Output a single integer — the number of ways in which Devu can select the flowers modulo (109?+?7).
Example
2 3
1 3
2
2 4
2 2
1
3 5
1 3 2
3
Note
Sample 1. There are two ways of selecting 3 flowers: {1,?2} and {0,?3}.
Sample 2. There is only one way of selecting 4 flowers: {2,?2}.
Sample 3. There are three ways of selecting 5 flowers: {1,?2,?2}, {0,?3,?2}, and {1,?3,?1}.
#include<iostream> #include<cstdio> #include<cstring> #define ll long long #define mod 1000000007 using namespace std; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } ll qpow(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 getc(ll a,ll b) { if(a<b)return 0; if(b>a-b)b=a-b; ll s1=1,s2=1; for(ll i=0;i<b;i++) { s1=s1*(a-i)%mod; s2=s2*(i+1)%mod; } return s1*qpow(s2,mod-2)%mod; } ll lucas(ll n,ll k) { if(k==0)return 1; return getc(n%mod,k%mod)*lucas(n/mod,k/mod)%mod; } int n; ll s,f[25]; ll solve() { ll ans=0; for(int i=0;i<(1<<n);i++) { ll sign=1,sum=s; for(int j=0;j<n;j++) { if(i&(1<<j)) { sum-=f[j]+1; sign*=-1; } } if(sum<0)continue; ans+=sign*lucas(sum+n-1,n-1); ans%=mod; } return (ans+mod)%mod; } int main() { cin>>n>>s; for(int i=0;i<n;i++) cin>>f[i]; printf("%lld\n",solve()); return 0; }
标签：line any sig ast int which lin UI blog
原文地址：http://www.cnblogs.com/joeylee97/p/7426659.html
