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J. Ceizenpok’s formula time limit per test ?2 seconds memory limit per test ?256 megabytes input ?standard input output ?standard output

Dr. Ceizenp'ok from planet i1c5l became famous across the whole Universe thanks to his recent discovery?— the Ceizenpok’s formula. This formula has only three arguments:?n,?k?and?m, and its value is a number of?k-combinations of a set of?n?modulo?m.

While the whole Universe is trying to guess what the formula is useful for, we need to automate its calculation.

Input

Single line contains three integers?n,?k,?m, separated with spaces (1?≤?n?≤?1018,?0?≤?k?≤?n,?2?≤?m?≤?1?000?000).

Output

Write the formula value for given arguments?n,?k,?m.

Sample test(s) input 2 1 3 output 2 input 4 2 5 output 1

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/* 2015 ICL, Finals, Div. 1 Ceizenpok’s formula（組合數取模，擴展lucas定理）求C(n,k)%m 如果m是素數的話直接就能套lucas模板. 對于m為合數,我們可以把它分解成素數在進行處理 m = p1*p2*p3..pk (pk = prime[i]^t) 然后利用擴展lucas定理可以求出 C(n,k) % pi的值，最后利用中國剩余定理漲姿勢：http://www.cnblogs.com/jianglangcaijin/p/3446839.html 題目鏈接：http://codeforces.com/gym/100633/problem/J hhh-2016-04-16 13:07:05 */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <functional> typedef long long ll; #define lson (i<<1) #define rson ((i<<1)|1) using namespace std;const int maxn = 1e6+10; ll fac[maxn]; int w[maxn],num[maxn],tw[maxn]; int tot; void get_factor(ll m) {ll mm = m;tot = 0;for(ll i = 2; i*i <= m; i++){if(mm % i == 0){num[tot] = 0;w[tot] = i;tw[tot] = 1;while(mm % i == 0){num[tot]++;mm /= i;tw[tot] *= i;}tot++;}}if(mm > 1){num[tot] = 1;w[tot] = mm;tw[tot] = mm;tot ++;} }ll ex_gcd(ll a,ll b,ll &x,ll &y) {if(a == 0 && b == 0)return -1;if(b == 0){x = 1,y = 0;return a;}ll d = ex_gcd(b,a%b,y,x);y -= a/b*x;return d; }ll pow_mod(ll a,ll b,ll mod) {ll ret = 1;while(b){if(b&1) ret = ret*a%mod;a = a*a%mod;b >>= 1;}return ret; }ll revers(ll a,ll b) {ll x,y;ll d = ex_gcd(a,b,x,y);if(d == 1) return (x%b+b)%b;else return 0; }ll c1(ll n,ll p,ll pk) {if(n==0)return 1;ll ans=1;for(ll i = 2; i <= pk; i++)if(i % p)ans = ans*i%pk;ans=pow_mod(ans,n/pk,pk);for(ll k=n%pk,i=2; i<=k; i++)if(i%p)ans=ans*i%pk;return ans*c1(n/p,p,pk)%pk; }ll cal(ll n,ll m,int cur,ll mod) {ll pi = w[cur],pk = tw[cur];ll k = 0,ans;ll a,b,c;a=c1(n,pi,pk),b=c1(m,pi,pk),c=c1(n-m,pi,pk);for(ll i=n; i; i/=pi)k+=i/pi;for(ll i=m; i; i/=pi)k-=i/pi;for(ll i=n-m; i; i/=pi)k-=i/pi;ans = a*revers(b,pk)%pk*revers(c,pk)%pk*pow_mod(pi,k,pk)%pk;return ans*(mod/pk)%mod*revers(mod/pk,pk)%mod; }ll lucas(ll n,ll m,ll mod) {ll ans = 0;for(int i = 0; i < tot; i++){ans = (ans+cal(n,m,i,mod))%mod;}return ans; }ll n,k; ll m; int main() {int T;while(scanf("%I64d%I64d%I64d",&n,&k,&m) != EOF){get_factor(m);printf("%I64d\n",lucas(n,k,m));}return 0; }

　　

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轉載于:https://www.cnblogs.com/Przz/p/5409567.html

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