Lucas theorem is used to find C (n, m) mod p, p is a prime number value. (Note: p must be a prime number)
Gives the number of combinations of C (n, m), m represents a selected number of program elements from the n elements. For example C (5,2) = 10, C (4,2) = 6. But when n, m when larger, C (n, m) great! So you want xiaobo output C (n, m) mod value of p!
Input
The first data input line T is a positive integer, the number (T <= 100) indicating that the following data set is a data group T, each data has three positive integers n, m, p (1 <= m <= n < = 10 ^ 9, m <= 10 ^ 4, m <p <10 ^ 9, p is a prime)
Output
For each test, it outputs a positive integer representing the C (n, m) mod p the result.
Sample Input
2 5 2 3 5 2 61
Sample Output
1 10
#include <cstdio> #define ll long long using namespace std; ll mulit(ll a,ll b,ll m) { ll ans=0; while(b) { if(b&1) { ans=(ans+a)%m; } a=(a<<1)%m; b>>=1; } return ans; } ll quick_mod(ll a,ll b,ll m) { ll ans=1; while(b) { if(b&1) { ans=mulit(ans,a,m); } a=mulit(a,a,m); b>>=1; } return ans; } ll comp(ll a,ll b,ll m) { if(a<b) { return 0; } if(a==b) { return 1; } if(b>a-b) { b=a-b; } ll ans=1,ca=1,cb=1; for(int i=0;i<b;i++) { ca=ca*(a-i)%m; cb=cb*(b-i)%m; } ans=ca*quick_mod(cb,m-2,m)%m; return ans; } ll lucas(ll a,ll b,ll m) { ll ans=1; while(a&&b) { ans=(ans*comp(a%m,b%m,m))%m; a/=m; b/=m; } return ans; } int main() { int T; ll n,m,p; scanf("%d",&T); while(T--) { scanf("%lld %lld %lld",&n,&m,&p); ll ans=lucas(n,m,p); printf("%lld\n",ans); } return 0; }
template
A, B are non-negative integers, p is a prime number. AB p hex written: A = a [n] a [n-1] ... a [0], B = b [n] b [n-1] ... b [0].
Number of combinations C (A, B) and C ([n] a, b [n]) * C (a [n-1], b [n-1]) * ... * C (a [0] , b [0]) modp congruence
即：Lucas(n,m,p)=c(n%p,m%p)*Lucas(n/p,m/p,p)
To solve for n! P%, for example, the n segments, each of the p section, each section is the same as a result of seeking. But need to be addressed separately at the end of each segment p, 2p, ..., p to extract, you will find just the remaining number is (n / p) !, became the equivalent of a child under the question, so recursive solution It can be.