2015ACM长春赛区网络赛 J题

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2049    Accepted Submission(s): 752

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

利用卢卡斯定理和中国剩余定理

#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;}