hdu5446 Unknown Treasure（Lucas+中国剩余定理）

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 pickm 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≤1018,1≤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≤1018 and pi≤105 for every i∈{1,...,k}.

Output

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

Sample Input

19 5 23 5

Sample Output

6

Source

2015 ACM/ICPC Asia Regional Changchun Online 

题意：M=p1*p2*...pk；求C（n，m）%M，pi小于10^5，n,m,M都是小于10^18。

分析：Lucas定理求C（n，m）%M，而中国剩余定理刚好解决后面部分。利用Lucas定理求出所有对pi取模的值，然后在用中国剩余定理求解。

Lucas定理不懂的可以戳这里；

中国剩余定理不懂的可以戳这里；

扩展欧几里德算法的话可以戳这里。

Ps. 刚开始学数论，这个题我看了三天博客，才略懂略懂，只能说数论这东西真尼玛难 - -

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 0x3f3f3f3f;const int MOD = 1000000007;#define ll long long#define CL(a) memset(a,0,sizeof(a))const int maxn = 1e5 + 5;ll fac[maxn]={0,1},inv[maxn]={0,1};ll pow_mod(ll a, ll n, ll mod)//快速幂取模{    ll ret = 1;    while (n)    {        if (n&1)            ret = ret*a%mod;        a = a*a%mod;        n >>= 1;    }    return ret;}ll Lucas(ll n, ll m, ll mod)//Lucas定理{    ll ret=1;    ll mm=m,nn=n;    while (mm && nn)    {        ll mod_m=mm%mod, mod_n=nn%mod;        if (mod_n>=mod_m)            ret = ret*fac[mod_n]%mod*inv[mod_m]%mod*inv[mod_n-mod_m]%mod;        else        {            ret = 0;            break;        }        mm/=mod;        nn/=mod;    }    return ret;}void exgcd(ll a, ll b, ll &d, ll &x, ll &y)//扩展欧几里德求逆元{    if (b == 0)        d = a, x = 1, y = 0;    else    {        exgcd(b, a%b, d, y, x);        y -= x * (a / b);    }}ll china(ll n, ll m[], ll a[])//中国剩余定理求解{    ll aa = a[0];    ll mm = m[0];    for (int i=0; i<n; i++)    {        ll sub = (a[i] - aa);        ll d, x, y;        exgcd(mm, m[i], d, x, y);        if (sub % d) return -1;        ll new_m = m[i]/d;        new_m = (sub/d*x%new_m+new_m)%new_m;        aa = mm*new_m+aa;        mm = mm*m[i]/d;    }    aa = (aa+mm)%mm;    return aa;}int main (){    int cas;    ll n, m, k, a[15], p[15];    scanf("%d", &cas);    while (cas--)    {        scanf("%lld%lld%lld",&n,&m,&k);        for (int i=0; i<k; i++)        {            scanf("%lld",&p[i]);            //init(p[i]);            for (int j=2; j<p[i]; j++)                fac[j] = fac[j-1]*j%p[i];            inv[p[i]-1] = pow_mod(fac[p[i]-1], p[i]-2, p[i]);            for (int j=p[i]-1; j>0; j--)                inv[j-1] = inv[j]*j%p[i];            a[i] = Lucas(n, m, p[i]);        }        printf("%lld\n",china(k, p, a));    }    return 0;}