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

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 HDU 5446 Unknown Treasure

Accept: 0    Submit: 0
Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

 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(T≤20) representing the number of test cases.

Each test case starts with three integers n,m,k(1≤m≤n≤10^18,1≤k≤10) 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=p1⋅p2⋅⋅⋅pk≤10^18 and pi≤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

 Problem Idea

解题思路:

【题意】
求解,M为k个不同质数的乘积

【类型】
lucas定理+中国剩余定理

【分析】
由于M为k个不同质数的乘积

所以M不一定为质数,这就导致我们无法直接使用lucas定理来求解组合数取模

那我们该怎么办呢?

先来回顾一下中国剩余定理

我们知道,对于一元模线性方程组

它的解可以写成x=a+kb(k≥0)的形式,并且

再回过头来看这道题,

在lucas定理不能直接使用的情况下,我们是否能够缩小,使得模运算可以直接计算呢?

答案显然是可以的

我们可以通过中国剩余定理,将缩小为等价的一元模线性方程组的最小解,这样就可以直接取模了

所以题目到此就转化为了求一元模线性方程组的最小解,即a

那么等价的一元模线性方程组是多少呢?如下:

于是乎,此题到此结束

【时间复杂度&&优化】

题目链接→HDU 5446 Unknown Treasure

 Source Code

/*Sherlock and Watson and Adler*/#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<bitset>#include<cmath>#include<complex>#include<string>#include<algorithm>#include<iostream>#define eps 1e-9#define LL long long#define PI acos(-1.0)#define bitnum(a) __builtin_popcount(a)using namespace std;const int N = 15;const int M = 100005;const int inf = 1000000007;const int mod = 7;__int64 quick_mod(__int64 a,__int64 b,__int64 p){    __int64 ans=1;    a%=p;    while(b)    {        if(b&1)        {            ans=ans*a%p;            b--;        }        b>>=1;        a=a*a%p;    }    return ans;}__int64 C(__int64 n,__int64 m,__int64 p){    if(m>n) return 0;    __int64 ans=1;    for(int i=1;i<=m;i++)    {        __int64 a=(n+i-m)%p;        __int64 b=i%p;        ans=ans*(a*quick_mod(b,p-2,p)%p)%p;    }    return ans;}__int64 Lucas(__int64 n,__int64 m,__int64 p){    if(m==0) return 1;    return C(n%p,m%p,p)*Lucas(n/p,m/p,p)%p;}int group;LL n[N],a[N];LL Egcd(LL a,LL b,LL &x,LL &y){    if(b==0)    {        x=1,y=0;        return a;    }    LL d,tp;    d=Egcd(b,a%b,x,y);    tp=x;    x=y;    y=tp-a/b*y;    return d;}LL solve(){    int i;    bool flag = false;    LL n1 = n[0], n2, b1 = a[0], b2, bb, d, t, k, x, y;    for (i = 1; i < group; i++)    {        n2 = n[i], b2 = a[i];        bb = b2 - b1;        d = Egcd (n1, n2, x, y);        if (bb % d)     //模线性解k1时发现无解        {            flag = true;            break;        }        k = bb / d * x;    //相当于求上面所说的k1【模线性方程】        t = n2 / d;        if (t < 0) t = -t;        k = (k % t + t) % t;    //相当于求上面的K`        b1 = b1 + n1*k;        n1 = n1 / d * n2;    }    if(flag)        return -1;     //无解/******************求正整数解******************/    if(b1==0)    //如果解为0，而题目要正整数解，显然不行        b1=n1;    //n1刚好为所有ni的最小公倍数，就是解了/******************求正整数解******************/    return b1;    //形成的解：b1, b1+n1, b1+2n1,..., b1+xni...}int main(){    int i,t;    __int64 p,q,M;    scanf("%d",&t);    while(t--)    {        M=1;        scanf("%I64d%I64d%d",&p,&q,&group);        for(i=0;i<group;i++)        {            scanf("%I64d",&n[i]);            M*=n[i];            a[i]=Lucas(p,q,n[i]);        }        printf ("%I64d\n",solve()%M);    }    return 0;}

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