GYM100633J. Ceizenpok’s formula 扩展lucas模板

J. Ceizenpok’s formula

time limit per test

2.0 s

memory limit per test

256 MB

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.

Examples

Input

2 1 3

Output

2

Input

4 2 5

Output

1

 

 

 

【题解】

裸的扩展lucas + CRT,推荐博文http://blog.csdn.net/clove_unique/article/details/54571216

因为特殊情况k = 0的时候，我以为是0，加了个特判，结果应该是1，不用特判。。。。。。。。。。卡了我半个晚上

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <string> 
#include <cmath> 
#include <sstream>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
template<class T>
inline void swap(T &a, T &b)
{
    T tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}
const long long INF = 0x3f3f3f3f;

long long pow(long long a, long long b, long long mod)
{
    long long r = 1, base = a % mod;
    for(;b;b >>= 1)
    {
        if(b & 1) r *= base, r %= mod;
        base *= base, base %= mod;
    }
    return r; 
}
void exgcd(long long a, long long b, long long &x, long long &y)
{
    if(!b)x = 1, y = 0;
    else exgcd(b, a%b, y, x), y -= (a / b) * x;
}
long long ni(long long x, long long mod)
{
    long long inv, y; exgcd(x, mod, inv, y);
    inv = (inv % mod + mod) % mod;
    if(!inv) inv = mod;
    return inv; 
}
int tiaoshi;
long long calc(long long n, long long p, long long pt)
{
    if(n == 0) return 1;
    long long ans = 1;
    for(long long i = 1;i <= pt;++ i) if(i % p) ans *= i, ans %= pt;
    ans = pow(ans, n/pt, pt);
    for(long long i = 1;i <= n%pt;++ i) 
        if(i % p) 
            ans *= i, ans %= pt;
    return ans * calc(n/p, p, pt) % pt;
}
long long C(long long n, long long m, long long p, long long pt)
{
    if(n < m || n < 0 || m < 0) return 0;
    long long cnt = 0;
    for(long long i = n;i;i /= p) cnt += i/p;
    for(long long i = m;i;i /= p) cnt -= i/p;
    for(long long i = n - m;i;i /= p) cnt -= i/p;
    return pow(p, cnt, pt) * calc(n, p, pt) % pt * ni(calc(m, p, pt), pt) % pt * ni(calc(n - m, p, pt), pt) % pt;
} 
long long exlucas(long long n, long long m, long long mod)
{
    long long tmp2 = mod, ans = 0;
    for(long long i = 2;i <= mod;++ i)
        if(tmp2 % i == 0)
        {
            long long pt = 1;
            while(tmp2 % i == 0)  tmp2 /= i, pt *= i;
            long long tmp3 = C(n, m, i, pt);
            ans += tmp3 * (mod/pt) % mod * ni(mod/pt, pt) % mod;
            ans %= mod;
        }
    return ans;
}
long long n,m,p;
int main()
{
    read(n), read(m), read(p);
    printf("%lld", exlucas(n, m, p)); 
    return 0;
}