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Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2610 Accepted Submission(s): 966
2 1 2 5 2 1 5
3 3HintHint For sample 1, squirrels will put no more than 2 beans in one tree. Since trees are different, we can label them as 1, 2 … and so on. The 3 ways are: put no beans, put 1 bean in tree 1 and put 2 beans in tree 1. For sample 2, the 3 ways are: put no beans, put 1 bean in tree 1 and put 1 bean in tree 2.
#include<map> #include<string> #include<cstring> #include<cstdio> #include<cstdlib> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<climits> #include<list> #include<iomanip> #include<stack> #include<set> using namespace std; typedef long long ll; ll qpow(ll a,ll b,ll c) { ll ans=1; while(b>0) { if(b&1) ans=ans*a%c; a=a*a%c; b>>=1; } return ans; } ll fac[100000]; ll fmt(ll a,ll b,ll c) { return fac[a]*qpow(fac[b]*fac[a-b]%c,c-2,c)%c; } ll lucas(ll a,ll b,ll c) { ll ans=1; while(a>0&&b>0) { if(a%c<b%c) return 0; ans=ans*fmt(a%c,b%c,c)%c; a/=c; b/=c; } return ans; } void create(ll a) { fac[0]=1; for(ll i=1;i<a;i++) fac[i]=fac[i-1]*i%a; } int main() { int T; cin>>T; while(T--) { ll n,m,p; cin>>n>>m>>p; create(p); cout<<lucas(n+m,m,p)<<endl; } }
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原文地址:http://www.cnblogs.com/mengfanrong/p/5083439.html