lightoj 1067 - Combinations （组合数、乘法逆元）

题目链接：http://lightoj.com/volume_showproblem.php?problem=1067

1067 - Combinations

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Given n different objects, you want to take k of them. How many ways to can do it?

For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways.

Take 1, 2

Take 1, 3

Take 1, 4

Take 2, 3

Take 2, 4

Take 3, 4

Input

Input starts with an integer T (≤ 2000), denoting the number of test cases.

Each test case contains two integers n (1 ≤ n ≤ 10⁶), k (0 ≤ k ≤ n).

Output

For each case, output the case number and the desired value. Since the result can be very large, you have to print the result modulo 1000003.

Sample Input	Output for Sample Input
3 4 2 5 0 6 4	Case 1: 6 Case 2: 1 Case 3: 15

 

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PROBLEM SETTER: JANE ALAM JAN

题目大意：就是求组合数  C (m,  n) 

解析：   由于数据比较多，用普通方法，先除再乘，运用循环，会 TLE 或者 WA， 会爆 long long， 所以打表就很好的解决了这个问题，

               打一个 1e6 的阶乘表， 由于 % 1000003.， 所以不会爆，然后就是运用 C (m,  n)  =   n !   /  (n - m ) !    /   m !  ，又因为是除法，

               故运用乘法逆元 （不知道的， 自己百度下）去解

代码如下：

#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#define N  1000009
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1000003;
long long dp[N];
void e_gcd(long long a, long long b, long long &x, long long &y)
{
    if(b == 0)
    {
        x = 1; y = 0;
        return ;
    }
    e_gcd(b, a% b, x, y);
    int tm = x; x = y;
    y = tm - a / b * y;
}

int main()
{
    int t, cnt = 0;
    long long m, n, ans, i, k, x, y;
    dp[0] = dp[1] = 1;
    for(i = 2; i <= 1000000; i++)
    {
        dp[i] = (dp[i - 1] * i) % mod;
        dp[i] %= mod;
    }
    cin >> t;
    while(t--)
    {
        scanf("%lld%lld", &n, &m);
        if(m == n || m == 0)
        {
            printf("Case %d: %lld\n", ++cnt, 1);
            continue;
        }
        e_gcd(dp[m], mod, x, y);
        x = ((x % mod) + mod) % mod;
        ans = x * dp[n] % mod;

        e_gcd(dp[n - m], mod, x, y);
        x = ((x % mod) + mod) % mod;
        ans = ans * x % mod;


        printf("Case %d: %lld\n", ++cnt, ans % mod);
    }
    return 0;
}