HDOJ Saving Beans 3037【Lucas定理】

Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2923    Accepted Submission(s): 1099

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. 
    
    
    
    
     
   
   
   
   

 

Source

2009 Multi-University Training Contest 13 - Host by HIT

 

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当要求的组合数比较大，而且对一个素数取模时。
普通的递推或者逆元的方法都不适用了。只能借助于Lucas定理。

百度百科——Lucas 定理：

数论Lucas定理是用来求 c(n,m) mod p的值,p是素数（从n取m组合，模上p）。
描述为:
          Lucas(n,m,p)=cm(n%p,m%p)* Lucas(n/p,m/p,p)
          Lucas(x,0,p)=1;
而
          cm(a,b)=a! * (b!*(a-b)!)^(p-2) mod p
           也= (a!/(a-b)!) * (b!)^(p-2)) mod p
这里，其实就是直接求 (a!/(a-b)!) / (b!) mod p
           由于 (a/b) mod p = a * b^(p-2) mod p

然后 插板法求排列组合 

题意中说不超过m个豆。

可以考虑m个的情况  并且可以有空  可以不放 所以可以考虑加上一棵树放不放的就变成 n+1 棵树。

等效为求：

  x1+x2+......+xn+1=m 的非负整数解个数。

因为非负的比较难求，所以我们可以转化成整数解。X1=X1+1，X2=X2+1......Xn=Xn+n;

即 X1+X2+.......+Xn+1=n+1+m
等效考虑就是：将n+1+m个球用n个隔板分成n+1份，这时一共有n+1+m-1=n+m个空
所以结果就是C(n,n+m)=C(m,n+m)

AC代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

using namespace std;

typedef long long LL;

LL f[100100];

void init(LL p)
{
    f[0]=1;
    for(int i=1;i<=p;i++)
        f[i]=f[i-1]*i%p;
}

long long mod_pow(LL x,LL n,LL p)
{
    LL res=1;
    while(n){
        if(n&1)res=res*x%p;
        x=x*x%p;
        n>>=1;
    }
    return res;
}


long long lucas(LL n,LL m,LL p)
{
    LL res=1;
    while(n&&m){
        int x=n%p;
        int y=m%p;
        if(x<y) return 0;
        if(y>x-y)y=x-y; //C（n,n+m）=c(m,n+m),选取较小值可以减少循环次数
        res=res*f[x]*mod_pow(f[y]*f[x-y]%p,p-2,p)%p;
        n/=p;m/=p;
    }
    return res;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        LL n,m,p;
        scanf("%lld%lld%lld",&n,&m,&p);
        init(p);
        printf("%lld\n",lucas(n+m,m,p));
    }
    return 0;
}