HDU 5446 Unknown Treasure（中国剩余定理+卢卡斯定理）——2015 ACM/ICPC Asia Regional Changchun Online

Posted 2022-04-16 ITAK

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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 $T (T \\leq 20)$ representing the number of test cases.

Each test case starts with three integers $n, m, k (1 \\leq m \\leq n \\leq 10^18, 1 \\leq k \\leq 10)$ on a line where $k$ is the number of primes. Following on the next line are $k$ different primes $p_1, . . . , p_k$. It is guaranteed that $M = p_1 · p_2 · · · p_k ≤ 10^18$ and $p_i \\leq 10^5$ 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
  

题目大意：

给定一个 $n,m(1\\le m\\le n\\le 10^18)$ ，求 $C(n,m)\\% M$ 其中 $M=p_1*p_2*...*p_k,其中p_i为素数$

解题思路：

设有 $A=C(n,m),X=A\\pmod M,Y=\\frac A M$，那么有

$$\\beginalign &A-Y*M=X\\\\ &A\\pmod p_i-Y*M\\pmod p_i=X\\pmod p_i\\\\ &A\\pmod p_i\\equiv X\\pmod p_i \\endalign$$
所以考虑中国剩余定理求得 $X$， 其中计算 $A\\pmod p_i$ 的时候用卢卡斯定理计算，最后需要用快速乘法，否则爆

long long

。

代码：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
using namespace std;
typedef long long LL;
const int MAXN = 1e5+5;
void Exgcd(LL a, LL b, LL &x, LL &y)
    if(b == 0)
        x = 1;
        y = 0;
        return;
    
    LL x1, y1;
    Exgcd(b, a%b, x1, y1);
    x = y1;
    y = x1 - (a/b)*y1;

LL Pow(LL a, LL b, LL c)

    LL ans = 1;
    while(b)
        if(b & 1) ans = (ans*a)%c;
        b>>=1;
        a = (a*a)%c;
    
    return ans;

LL Multi(LL a, LL b, LL c)
    LL ans = 0;
    while(b)
        if(b & 1) ans = (ans+a)%c;
        b>>=1;
        a = (a+a)%c;
    
    return ans;

int C(int a, int b, int c)
    if(b > a) return 0;
    int tmp = min(b, a-b);
    LL t1 = 1, t2 = 1;
    for(int i=1; i<=tmp; i++)
        t1 = t1*(a-i+1)%c;
        t2 = t2*i%c;
    
    LL x, y;
    Exgcd(t2, c, x, y);
    LL ans = t1*x%c;
    ans = (ans%c+c)%c;
    return ans;

LL Lucas(LL n, LL m, int p)
    LL ans = 1;
    while(n && m)
        ans = ans*C(n%p, m%p, p) % p;
        n /= p;
        m /= p;
    
    return ans;

LL p[15], a[15];
LL CRT(int n)
    LL M = 1, ans = 0;
    for(int i=0; i<n; i++) M *= p[i];
    for(int i=0; i<n; i++)
        LL _M = M/p[i];
        LL x, y;
        Exgcd(_M, p[i], x, y);
        x = (x%p[i]+p[i])%p[i];
        LL tmp = Multi(x, _M, M);
        ans = (ans+Multi(tmp, a[i], M))%M;
    
    ans = (ans%M+M)%M;
    return ans;

int main()
    int T; scanf("%d", &T);
    while(T--)
        LL n, m; scanf("%lld%lld",&n,&m);
        int k; scanf("%d",&k);
        for(int i=0; i<k; i++) scanf("%lld",&p[i]);
        for(int i=0; i<k; i++) a[i] = Lucas(n, m, p[i]);
        printf("%lld\\n",CRT(k));
    
    return 0;