hdoj 5446 Unknown Treasure 【lucas + CRT】



Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1820    Accepted Submission(s): 671


Problem Description
On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician entered the cave because it is there. Somewhere deep in the cave, she found a treasure chest with a combination lock and some numbers on it. After quite a research, the mathematician found out that the correct combination to the lock would be obtained by calculating how many ways are there to pick  m  different apples among  n  of them and modulo it with  M M  is the product of several different primes.
 

Input
On the first line there is an integer  T(T20)  representing the number of test cases.

Each test case starts with three integers  n,m,k(1mn1018,1k10)  on a line where  k  is the number of primes. Following on the next line are  k different primes  p1,...,pk . It is guaranteed that  M=p1p2pk1018  and  pi105  for every  i{1,...,k} .
 

Output
For each test case output the correct combination on a line.
 

Sample Input
       
       
       
       
1 9 5 2 3 5
 

Sample Output
       
       
       
       
6
 



题意:求解C(n, m) % (p[0] * p[1] * p[2] * ... * p[k-1])。 其中p[]是不同的质数。


思路:由于p[0] * p[1] * ... * p[k-1]很大,所以不不能直接用lucas。

考虑把每个质数分开。

C(n, m) % p[0] = a[0], C(n, m) % p[1] = a[1]... C(n, m) % p[k-1] = a[k-1]。

这样得到下面方程组

A = a[0] (mod p[0])

A = a[1] (mod p[1])

...

A = a[k-1] (mod p[k-1])。

下面用CRT直接求出A就可以了。

因为题目没有说结果是正整数,所以CRT最后解为0时,直接返回。这里WA了3次 o(╯□╰)o



AC代码:


#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define debug printf("1\n");
using namespace std;
struct ONE
{
    LL gcd(LL a, LL b){
        return b == 0 ? a : gcd(b, a%b);
    }
    LL lcm(LL a, LL b){
        return a / gcd(a, b) * b;
    }
    void exgcd(LL a, LL b, LL &d, LL &x, LL &y)
    {
        if(b == 0) {d = a, x = 1, y = 0;}
        else
        {
            exgcd(b, a%b, d, y, x);
            y -= x * (a / b);
        }
    }
    LL CRT(LL l, LL r, LL *a, LL *m)
    {
        LL LCM = 1;
        for(LL i = l; i <= r; i++)
            LCM = LCM / gcd(LCM, m[i]) * m[i];
        for(LL i = l+1; i <= r; i++)
        {
            LL d, x, y, A = m[l], B = m[i], c = a[i] - a[l];
            exgcd(A, B, d, x, y);
            if(c % d) return -1;
            LL mod = m[i] / d;
            LL k = ((x / d * c) % mod + mod) % mod;
            a[l] = m[l] * k + a[l];
            m[l] = m[l] / d * m[i];
        }
        //if(a[l] == 0) 这句话不能加
          //  return LCM;
        return a[l];
    }
};
ONE crt;
LL fac[100000+10];
void getfac(LL p)
{
    fac[0] = 1 % p;
    for(LL i = 1; i <= p; i++)
        fac[i] = fac[i-1] * i % p;
}
struct TWO
{
    LL pow_mod(LL a, LL n, LL p)
    {
        LL ans = 1 % p;
        while(n)
        {
            if(n & 1)
                ans = ans * a % p;
            a = a * a % p;
            n >>= 1;
        }
        return ans;
    }
    LL C(LL n, LL m, LL p)
    {
        if(n < m)
            return 0;
        else
            return fac[n] * pow_mod(fac[m]*fac[n-m]%p, p-2, p) % p;
    }
    LL Lucas(LL n, LL m, LL p)
    {
        if(m == 0)
            return 1 % p;
        else
            return C(n%p, m%p, p) * Lucas(n/p, m/p, p) % p;
    }
};
TWO LUCAS;
LL p[15], a[15], m[15];
int main()
{
    LL t;
    scanf("%d", &t);
    while(t--)
    {
        LL N, M, k;
        scanf("%lld%lld%lld", &N, &M, &k);
        for(LL i = 0; i < k; i++)
        {
            scanf("%lld", &p[i]);
            m[i] = p[i];
            getfac(p[i]);
            a[i] = LUCAS.Lucas(N, M, p[i]);
        }
        printf("%lld\n", crt.CRT(0, k-1, a, p));
    }
    return 0;
}



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