BZOJ2982: combination Lucas模板

2982: combination

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 734  Solved: 437
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Description

LMZ有n个不同的基友，他每天晚上要选m个进行[河蟹]，而且要求每天晚上的选择都不一样。那么LMZ能够持续多少个这样的夜晚呢？当然，LMZ的一年有10007天，所以他想知道答案mod 10007的值。(1<=m<=n<=200,000,000)

Input

  第一行一个整数t，表示有t组数据。(t<=200)

  接下来t行每行两个整数n, m，如题意。

Output

T行，每行一个数，为C(n, m) mod 10007的答案。

Sample Input

4
5 1
5 2
7 3
4 2

Sample Output

5
10
35
6

HINT

Source

lucas模板题

#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;
const long long MAXN = 200000000;
const long long MOD = 10007;

long long f[MOD + 1000], t, n, m;
long long pow(long long a, long long b)
{
    long long r = 1, base = a % MOD;
    for(;b;b >>= 1)
    {
        if(b & 1) r *= base, r %= MOD;
        base *= base, base %= MOD;
    }
    return r;
}
long long ni(long long x)
{
    return pow(x, MOD - 2);
}
long long C(long long n, long long m)
{
    if(n < m) return 0;
    int tmp = f[n] * ni(f[m]) % MOD , tmp2 = tmp * ni(f[n - m]) % MOD;
    return tmp2;
}
long long lucas(long long n, long long m)
{
    if(!m) return 1;
    else return C(n % MOD, m % MOD) * lucas(n / MOD, m / MOD) % MOD;
}

int main()
{
    read(t);f[0] = 1;
    for(long long i = 1;i <= MOD;++ i)
        f[i] = f[i - 1] * i % MOD;
    for(long long i = 1;i <= t;++ i)
    {
        read(n), read(m);
        printf("%lld
", lucas(n, m));
    }
    return 0;
}