HDU 3037 Saving Beans【Lucas定理】【模板题】【模板】【组合数取余】

Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5775    Accepted Submission(s): 2320

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.

A、B是非负整数，p是质数。AB写成p进制：A=a[n]a[n-1]...a[0]，B=b[n]b[n-1]...b[0]。
则组合数C(A,B)与C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0])  mod p同余

即：Lucas(n，m，p)=c(n%p，m%p)*Lucas(n/p，m/p，p) 

Lucas最大的数据处理能力是p在10^5左右，不能再大了，hdu 3037就是10^5级别的！

对于大组合数取模，n,m不大于10^5的话，用逆元的方法，可以解决。对于n,m大于10^5的话，那么要求p<10^5，这样就是Lucas定理了，将n,m转化到10^5以内解。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define ms(x,y) memset(x,y,sizeof(x))
using namespace std;

typedef long long ll;

const int maxn = 1e5 + 5;

ll fac[maxn];

void init(ll mod)	//初始化乘积，优化C(n,m,mod)
{
    ll i;
    fac[0] =1;
    for(i =1; i <= mod; i++)
        fac[i] = fac[i-1]*i % mod;
}

ll exp_mod(ll a, ll b, ll mod)	//快速幂
{
    ll tmp = a % mod, ans =1;
    while(b)
    {
        if(b & 1)  ans = ans * tmp % mod;
        tmp = tmp*tmp % mod;
        b >>=1;
    }
    return  ans;
}

ll C(ll n, ll m, ll mod)	//朴素组合数n取m
{
    if(m > n)
    	return 0;
    return  fac[n]*exp_mod(fac[m]*fac[n-m], mod-2, mod) % mod;//逆元
}

ll Lucas(ll n, ll m, ll mod)	//lucas求组合数c n取m 
{
    if(m ==0)
    	return 1;
    return  (C(n%mod, m%mod, mod)*Lucas(n/mod, m/mod, mod))%mod;
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		ll n, m, mod;
		scanf("%I64d%I64d%I64d", &n, &m, &mod);
		init(mod);			//注意要初始化乘积
		//printf("%I64d
", C(n+m,m,mod));
		printf("%I64d
", Lucas(n + m, m, mod));
	}
	return 0;
}

朴素T

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define ms(x,y) memset(x,y,sizeof(x))
using namespace std;

typedef long long ll;

const int maxn = 1e5 + 5;

ll quick_pow(ll a, ll b, ll mod)        //快速幂
{
    ll ans = 1;
    a %= mod;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a % mod;
        }
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

ll C(ll n, ll m, ll mod)    //朴素求组合数n取m
{
    if (m > n) return 0;
    ll ans = 1;
    for (int i = 1; i <= m; i++)
    {
        ll a = (n + i - m) % mod;
        ll b = i % mod;
        ans = ans * (a * quick_pow(b, mod - 2, mod) % mod) % mod;    //逆元求组合数
    }
    return ans;
}

ll lucas(ll n, ll m, ll mod)    //Lucas求组合数n取m
{
    if (m == 0) return 1;
    return C(n % mod, m % mod, mod) * lucas(n / mod, m / mod, mod) % mod;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t;
    scanf("%d", &t);
    while (t--)
    {
        ll n, m, mod;
        scanf("%I64d%I64d%I64d", &n, &m, &mod);
        //printf("%I64d
", C(n+m,m,mod));
        printf("%I64d
", lucas(n + m, m, mod));
    }
    return 0;
}