[Lucas Theorem] [Method separator] [inclusion and exclusion] Codeforces 451E Devu and Flowers

 

 

 answer

• We first consider the case where there is no limit how demand, i.e. remove the n box number s balls embodiment, each box is a separator may not take Method C (n + s-1, n-1)
• Consider then subtract each illegal scheme, which is the number of i boxes take over the program
• Consider the inclusion-exclusion principle, it is easy to think of inclusion and exclusion factor (-1) ^ i, then took the total number of programs - 1 + 2 box box super super ...
• Then find or use Lucas theorem, n and m are as much

 

Code

 1 #include <cstdio>
 2 #include <iostream>
 3 #define ll long long
 4 using namespace std;
 5 const ll mo=1e9+7,N=30;
 6 int n;
 7 ll s,r,f[N];
 8 ll ksm(ll a,ll b) { for (r=1;b;b>>=1,a=a*a%mo) if (b&1) r=r*a%mo; return r; }
 9 ll C(ll n,ll m) 
10 { 
11     if (m>n) return 0;
12     if (m>n-m) m=n-m;
13     ll ans=1ll,fac=1ll;
14     for (int i=1;i<=m;i++) ans=(ans*(n-i+1))%mo,fac=fac*i%mo;
15     fac=ksm(fac,mo-2),ans=ans*fac%mo;
16     return ans;
17 }
18 ll lucas(ll n,ll m) { return !m?1:C(n%mo,m%mo)*lucas(n/mo,m/mo)%mo; }
19 int main()
20 {
21     while (scanf("%d%lld",&n,&s)!=EOF)
22     {
23         ll ans=0;
24         for (int i=0;i<n;i++) scanf("%lld",&f[i]);
25         for (int i=0;i<(1<<n);i++)
26         {
27             int k=1; ll p=s;
28             for (int j=0;j<n;j++) if ((1<<j)&i) k*=-1,p-=f[j]+1;
29             if (p<0) continue;
30             (ans+=k*lucas(p+n-1,n-1)%mo)%=mo;
31         }
32         ans=(ans+mo)%mo,printf("%lld\n",ans);
33     }
34 }