HDU 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): 4540    Accepted Submission(s): 1788

 

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

 

 

求n个数的和不超过m的方案数。

题目可以转换成  x1+x2+……+xn=m 有多少组解,m在题中可以取0~m。
利用插板法可以得出x1+x2+……+xn=m解的个数为C(n+m-1,m);
则题目解的个数可以转换成求 sum=C(n+m-1,0)+C(n+m-1,1)+C(n+m-1,2)……+C(n+m-1,m)
利用公式C(n,r)=C(n-1,r)+C(n-1,r-1)  == >  sum=C(n+m,m)。
现在就是要求C(n+m,m)%p。
因为n,m很大,这里可以直接套用Lucas定理的模板即可。


AC代码:

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

typedef __int64 LL;

LL exp_mod(LL a, LL b, LL p)
{
    LL res = 1;
    while(b != 0)
    {
        if(b&1) res = (res * a) % p;
        a = (a*a) % p;
        b >>= 1;
    }
    return res;
}

LL Comb(LL a, LL b, LL p)
{
    if(a < b)return 0;
    if(a == b)return 1;
    if(b > a - b)b = a - b;

    LL ans = 1, ca = 1, cb = 1;
    for(LL i = 0; i < b; ++i)
    {
        ca = (ca * (a - i))%p;
        cb = (cb * (b - i))%p;
    }
    ans = (ca*exp_mod(cb, p - 2, p)) % p;
    return ans;
}

LL Lucas(int n, int m, int p)
{
    LL ans = 1;

    while(n&&m&&ans)
    {
        ans = (ans*Comb(n%p, m%p, p)) % p;
        n /= p;
        m /= p;
    }
    return ans;
}

int main()
{
    int n, m, p,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d", &n, &m, &p);
        printf("%I64d\n", Lucas(n+m, m, p));    //C(n+m,m)%p
    }
    return 0;
}