HDOJ 3037 Saving Beans
如果您有n+1树,文章n+1埋不足一棵树m种子,法国隔C【n+m】【m】
大量的组合,以取mod使用Lucas定理:
Lucas(n,m,p) = C[n%p][m%p] × Lucas(n/p,m/p,p) ;
Saving Beans
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2314 Accepted Submission(s): 845
in n different trees. However, since the food is not sufficient nowadays, they will get no more than m beans. They want to know that how many ways there are to save no more than m beans (they are the same) in n trees.
Now they turn to you for help, you should give them the answer. The result may be extremely huge; you should output the result modulo p, because squirrels can’t recognize large numbers.
Then followed T lines, each line contains three integers n, m, p, means that squirrels will save no more than m same beans in n different trees, 1 <= n, m <= 1000000000, 1 < p < 100000 and p is guaranteed to be a prime.
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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; typedef long long int LL; LL n,m,p; LL fact[100100]; LL QuickPow(LL x,LL t,LL m)
{
if(t==0) return 1LL;
LL e=x,ret=1LL;
while(t)
{
if(t&1) ret=(ret*e)%m;
e=(e*e)%m;
t>>=1LL;
}
return ret%m;
} void get_fact(LL p)
{
fact[0]=1LL;
for(int i=1;i<=p+10;i++)
fact[i]=(fact[i-1]*i)%p;
} LL Lucas(LL n,LL m,LL p)
{
///lucas(n,m,p)=c[n%p][m%p]*lucas(n/p,m/p,p);
LL ret=1LL;
while(n&&m)
{
LL a=n%p,b=m%p;
if(a<b) return 0;
ret=(ret*fact[a]*QuickPow((fact[b]*fact[a-b])%p,p-2,p))%p;
n/=p; m/=p;
}
return ret%p;
} int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
LL n,m,p;
cin>>n>>m>>p;
get_fact(p);
cout<<Lucas(n+m,m,p)<<endl;
}
return 0;
}
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