Unknown Treasure
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2389 Accepted Submission(s): 885
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≤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} .
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
1
9 5 2
3 5
Sample Output
6
Source
题意：C(n,m)%p1*p2*p3..pk
题解：中国剩余定理+lucas
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #define ll __int64 6 #define mod 10000000007 7 using namespace std; 8 ll n,m,k; 9 int t; 10 ll exm; 11 ll f[1000010]; 12 void init(int p) { //f[n] = n! 13 f[0] = 1; 14 for (int i=1; i<=p; ++i) f[i] = f[i-1] * i % p; 15 } 16 ll mulmod(ll x,ll y,ll m) 17 { 18 ll ans=0; 19 while(y) 20 { 21 if(y%2) 22 { 23 ans+=x; 24 ans%=m; 25 } 26 x+=x; 27 x%=m; 28 y/=2; 29 } 30 ans=(ans+m)%m; 31 return ans; 32 } 33 34 void exgcd(ll a, ll b, ll &x, ll &y) 35 { 36 if(b == 0) 37 { 38 x = 1; 39 y = 0; 40 return; 41 } 42 exgcd(b, a % b, x, y); 43 ll tmp = x; 44 x = y; 45 y = tmp - (a / b) * y; 46 } 47 48 ll pow_mod(ll a, ll x, int p) { 49 ll ret = 1; 50 while (x) { 51 if (x & 1) ret = ret * a % p; 52 a = a * a % p; 53 x >>= 1; 54 } 55 return ret; 56 } 57 ll CRT(ll a[],ll m[],ll n) 58 { 59 ll M = 1; 60 ll ans = 0; 61 for(ll i=0; i<n; i++) 62 M *= m[i]; 63 for(ll i=0; i<n; i++) 64 { 65 ll x, y; 66 ll Mi = M / m[i]; 67 exgcd(Mi, m[i], x, y); 68 ans = (ans +mulmod( mulmod( x , Mi ,M ), a[i] , M ) ) % M; 69 } 70 ans=(ans + M )% M; 71 return ans; 72 } 73 74 ll Lucas(ll n, ll k, ll p) { //C (n, k) % p 75 init(p); 76 ll ret = 1; 77 while (n && k) { 78 ll nn = n % p, kk = k % p; 79 if (nn < kk) return 0; //inv (f[kk]) = f[kk] ^ (p - 2) % p 80 ret = ret * f[nn] * pow_mod (f[kk] * f[nn-kk] % p, p - 2, p) % p; 81 n /= p, k /= p; 82 } 83 return ret; 84 } 85 int main () 86 { 87 scanf("%d",&t); 88 { 89 for(int i=1;i<=t;i++) 90 { 91 ll ee[15]; 92 ll gg[15]; 93 scanf("%I64d %I64d %I64d",&n,&m,&k); 94 for(ll j=0;j<k;j++) 95 { 96 scanf("%I64d",&exm); 97 gg[j]=exm;; 98 ee[j]=Lucas(n,m,exm); 99 } 100 printf("%I64d ",CRT(ee,gg,k)); 101 } 102 } 103 return 0; 104 }