hdu 5446 Unknown Treasure 卢卡斯+中国剩余定理
Unknown Treasure
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 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 and pi≤105 for every i∈{1,...,k}.
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 and pi≤105 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
Source
题意:求c(n,m)%(p1*p2*...*pk);
思路:求出每个c(n,m)%p1=a1......求出a数组;
然后根据a求中国剩余即是答案;
#include<bits/stdc++.h> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-14 const int N=1e5+10,M=1e6+10,inf=1e9+10; const ll INF=1e18+10,mod=2147493647; ll p[20],a[20]; ll n,m; ll mulmod(ll x,ll y,ll m) { ll ans=0; while(y) { if(y%2) { ans+=x; ans%=m; } x+=x; x%=m; y/=2; } ans=(ans+m)%m; return ans; } ll ff(ll x,ll p) { ll ans=1; for(int i=1;i<=x;i++) ans*=i,ans%=p; return ans; } ll pow_mod(ll a, ll x, ll p) { ll ret = 1; while (x) { if (x & 1) ret = ret * a % p; a = a * a % p; x >>= 1; } return ret; } ll Lucas(ll n, ll k, ll p) { //C (n, k) % p ll ret = 1; while (n && k) { ll nn = n % p, kk = k % p; if (nn < kk) return 0; //inv (f[kk]) = f[kk] ^ (p - 2) % p ret = ret * ff(nn,p) * pow_mod (ff(kk,p) * ff(nn-kk,p) % p, p - 2, p) % p; n /= p, k /= p; } return ret; } void exgcd(ll a, ll b, ll &x, ll &y) { if(b == 0) { x = 1; y = 0; return; } exgcd(b, a % b, x, y); ll tmp = x; x = y; y = tmp - (a / b) * y; } ll CRT(ll a[],ll m[],ll n) { ll M = 1; ll ans = 0; for(ll i=1; i<=n; i++) M *= m[i]; for(ll i=1; i<=n; i++) { ll x, y; ll Mi = M / m[i]; exgcd(Mi, m[i], x, y); //ans = (ans + Mi * x * a[i]) % M; ans = (ans +mulmod( mulmod( x , Mi ,M ), a[i] , M ) ) % M; } ans=(ans + M )% M; return ans; } int main() { int T; scanf("%d",&T); while(T--) { int k; scanf("%lld%lld%d",&n,&m,&k); for(int i=1;i<=k;i++) scanf("%lld",&p[i]); for(int i=1;i<=k;i++) a[i]=Lucas(n,m,p[i]); printf("%lld\n",CRT(a,p,k)); } return 0; }