HDU3037 附Lucas简单整理

隔板法与Lucas定理解决组合计数问题

最新推荐文章于 2025-09-13 09:06:35 发布

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最新推荐文章于 2025-09-13 09:06:35 发布

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文章标签： java

原文链接：http://www.cnblogs.com/shenben/p/6529140.html

本文介绍了一种利用隔板法和Lucas定理解决组合计数问题的方法，该问题是在特定条件下计算放置球到不同盒子的方案数。通过实例解释了如何将大范围的组合数计算转换为模意义下的运算。

Saving Beans

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4941 Accepted Submission(s): 1957

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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题目大意：

由n个不同的盒子，在每个盒子中放一些求(可以不放)，使得总球数小雨等于m,求方案数(mod p).
1<=n,m<=10^9,1<p<10^5,保证p是素数

分析：
设最后放了k个球，根据"隔板法"由方案数C(k+n-1,n-1),:
ans=C(n-1,n-1)+C(n,n-1)+C(n+1,n-1)+……+C(n+m-2,n-1)+C(n+m-1,n-1)
=C(n+m,n);(mod p)
由于数据范围很大,C(n,m)=C(n-1,m)+C(n-1,m-1);显然会TLE
最后组合数还要mod p，这时候 Lucas定理闪亮登场
============================================
Lucas定理(shenben简单总结版)
============================================
Lucas定理1：
　　Lucas(n,m,p)=cm(n%p,m%p)*Lucas(n/p,m/p,p);{其中cm(a,b)=C(a,b)%p;Lucas(x,0,p)=1;}
Lucas定理2：
　　把n写成p进制a[n]a[n-1]……a[0];
　　把m写成p进制b[n]b[n-1]……b[0];(不够位数的话，显然前面是 0)
则：C(a[n],b[n])*C(a[n-1],b[n-1])*……*C(a[0],b[0])
=C(n,m) (mod p);
ps:Lucas最大的数据处理能力是p在10^5左右。

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
const int N=1e5+5;
ll n,m,p,a[N]={1};
ll fpow(ll a,ll b){
    ll res=1;
    for(;b;b>>=1,a=a*a%p) if(b&1) res=res*a%p;
    return res;
}
ll C(ll n,ll m){
    if(m>n) return 0;
    return a[n]*fpow(a[m],p-2)%p*fpow(a[n-m],p-2)%p;
}
ll lucas(ll n,ll m){
    if(!m) return 1;
    return C(n%p,m%p)*lucas(n/p,m/p)%p;
}
int main(){
    int T;cin>>T;
    while(T--){
        cin>>n>>m>>p;
        for(int i=1;i<=p;i++) a[i]=a[i-1]*i%p;
        cout<<lucas(n+m,n)<<'\n';
    }
    return 0;
}