HDOJ 3037 Saving Beans （Lucas定理）

Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3533    Accepted Submission(s): 1350

Problem Description

Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel family thinks that they have to solve a problem. They suppose that they will save beans 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.

 

Input

The first line contains one integer T, means the number of cases.

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.

 

Output

You should output the answer modulo p.

 

Sample Input

   
   
   
   

    
    
    
    2
1 2 5
2 1 5
   
   
   
   

 

Sample Output

  
  
  
  

   
   
   
   3
3

   
   
   
   
 
    
 
     Hint 
    
 Hint 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. 
     思路：简单的组合数学，ans=C(n+m,m)%P,因为涉及到取模而且数据较大，所以说就是Lucas定理了 ac代码：   
    
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
LL f[MAXN];
void findinit(LL p)
{
	f[0]=1;
	for(int i=1;i<=p;i++)
	f[i]=(f[i-1]*i)%p;
}
LL lucas(LL n,LL m,LL p)
{
	LL ans=1;
	while(n&&m)
	{
		LL a=n%p,b=m%p;
		if(a<b)
		return 0;
		ans=(ans*f[a]*powmod(f[b]*f[a-b]%p,p-2,p))%p;
		n/=p;m/=p;
	}
	return ans;
}
int main()
{
	LL n,m,p;
	int t;
	scanf("%d",&t); 
	while(t--)
	{
		scanf("%lld%lld%lld",&n,&m,&p);
		findinit(p);
		LL ans=lucas(n+m,m,p);
		printf("%lld\n",ans);
	}
	return 0;
}