LightOJ-1067 - Combinations

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

借鉴：https://blog.csdn.net/qq_34287501/article/details/52143890?locationNum=6

#include<iostream>
#include<string>
#include<cstdio>
#define N  1000009
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1000003;
long long dp[N];
long long ex_gcd(long long a,long long b,long long &x,long long &y)
{
    long long d;
    if(b==0)
    {x=1; y=0;  return a;}
    d=ex_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
int main()
{
    int t, cnt = 0;
    long long m, n, ans, i, x, y;
    dp[0] = dp[1] = 1;                      //C(n,m)公式：n!/(m!*(n-m)!)
    for(i = 2; i <= 1000000; i++)           //C(n,m)会爆数据什么的吧
    {                                       //模p乘法逆元的应用:将除法mod转为乘法
                                            // (a/b)%p==(a*b1)%p, b1为b的逆元
                                            // (a/(b*c))%p==(a*b1*c1)%p,b1为b的逆元,c1为c的逆元
        dp[i] = (dp[i - 1] * i) % mod;
        dp[i] %= mod;                      //阶乘mod的话，从1开始推的话就可以推出这个公式。
    }                                      //所以现在只要记住累乘mod两下就行
    cin >> t;                              //就叫做阶乘模吧。。。。
    while(t--)
    {
        scanf("%lld%lld", &n, &m);
        if(m == n || m == 0)               //C(m,0)==1;
        {
            printf("Case %d: %lld\n", ++cnt, 1);
            continue;
        }
        ex_gcd(dp[m], mod, x, y);
        x = ((x%mod)+mod)% mod;  //这样就保证是最小正数
        ans = x%mod;
        ex_gcd(dp[n - m], mod, x, y);
        x = ((x % mod) + mod) % mod;   //公式算的是一个因子的逆元，当两个因子相乘的话
        ans = ans*x%mod;
        ans%=mod;
        printf("Case %d: %lld\n", ++cnt, (ans* dp[n])% mod);
    }
    return 0;
}