BZOJ1272: [BeiJingWc2008]Gate Of Babylon

题解:

多重集合的组合数?还是0-m?有些元素有个数限制?

多重集合的组合数可以插板法,0-m直接利用组合数的公式一遍求出来,个数限制注意到只有15个,那我们就暴力容斥了

AC了真舒畅。。

注意开long long

 ll n,m,a[],k,p,ans,fac[maxn],inv[maxn];
inline ll c(ll n,ll m)
{
if(n<m)return ;
if(n<p&&m<p)return fac[n]*inv[m]%p*inv[n-m]%p;
return c(n%p,m%p)*c(n/p,m/p)%p;
}
inline void dfs(int x,int w1,ll w2)
{
if(x==k+){ans=(ans+w1*(c(m+n-w2,n)-c(n-w2-,n)))%p;return;}//cout<<n<<' '<<m<<' '<<w2<<' '<<n+m-w2<<' '<<n<<' '<<ans<<endl;return;}
dfs(x+,-w1,w2+a[x]+);
dfs(x+,w1,w2);
} int main() { freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); n=read();k=read();m=read();p=read();
fac[]=;
for1(i,p-)fac[i]=fac[i-]*i%p;
inv[]=inv[]=;
for2(i,,p-)inv[i]=inv[i-p%i]*(p/i+)%p;
for1(i,p-)inv[i]=inv[i]*inv[i-]%p;
for1(i,k)a[i]=read();
dfs(,,);
cout<<(ans+p)%p<<endl; return ; }

1272: [BeiJingWc2008]Gate Of Babylon

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 195  Solved: 90
[Submit][Status]

Description

BZOJ1272: [BeiJingWc2008]Gate Of Babylon

Input

BZOJ1272: [BeiJingWc2008]Gate Of Babylon

Output

BZOJ1272: [BeiJingWc2008]Gate Of Babylon

Sample Input

BZOJ1272: [BeiJingWc2008]Gate Of Babylon

Sample Output

12
BZOJ1272: [BeiJingWc2008]Gate Of Babylon

HINT

BZOJ1272: [BeiJingWc2008]Gate Of Babylon

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