Saving Beans(hud3037)

Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4153    Accepted Submission(s): 1607

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：隔板法+lucas定理；

 题意：在n个树上放小于m个苹果有多少种方案；

 首先我们先考虑放m个苹果在n棵树上有多少种方案，问题转化为求x1+x2+...xn=m的方案数。

 这个就可以用隔板法来求，那么就是C(n+m-1,n-1)=C(n+m-1,m);

 那么答案就是C（n-1,0）+C(n,1)+C(n+1,2)+...C(n+m-1,m);

 根据杨辉三角上式等于C（n,0）+C(n,1)+C(n+1,2)+...C(n+m-1，m)；逐项两两合并就可以得到C（n+m,m）；

 那么由于primep比较小，并且n+m比较大，所以用lucas定理去求；

 

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<map>
#include<math.h>
using namespace std;
typedef long long LL;
LL quick(LL n,LL m,LL mod);
LL lucas(LL n,LL m,LL mod);
LL a[100005];
int main(void)
{
        int i,j,k;
        LL x,y,z;
        scanf("%d",&k);
        while(k--)
        {
                scanf("%lld %lld %lld",&x,&y,&z);
                a[0]=1;
                a[1]=1;
                for(i=2; i<=z; i++)
                {
                        a[i]=a[i-1]*i;
                        a[i]%=z;
                }
                LL n=(x+y);
                LL m=x;
                LL ask=lucas(m,n,z);
                printf("%lld
",ask%z);
        }
        return 0;
}
LL quick(LL n,LL m,LL mod)
{
        LL ans=1;
        while(m)
        {
                if(m&1)
                {
                        ans=ans*n%mod;
                }
                n=n*n%mod;
                m/=2;
        }
        return ans;
}
LL lucas(LL n,LL m,LL mod)
{
        if(n==0)
        {
                return 1;
        }
        else
        {
                LL x1=n/mod;
                LL x2=m/mod;
                LL t1=n%mod;
                LL t2=m%mod;
                LL t3=a[t2-t1]*a[t1]%mod;
                if(t2<t1)return 0;
                LL nit3=quick(t3,mod-2,mod);
                return  (nit3*a[t2]%mod*lucas(x1,x2,mod)%mod)%mod;
        }
}