light oj 1067 组合数取模

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.

题目连接：http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=26784

题目：

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

题意：给你n和k，从n个different objects，选k个不同的，就是求组合数C（n,k）%1000003；

分析：

时间只有2秒，T组测试数据加上n的106达到了109递推肯定超时，那么考虑组合公式，C(n,k)=n!/(k!*(n-k)!);先打一个阶乘的表（当然要取模，只有106）,然后就是这个除法取模的问题，当然是求逆元，（a/b）%mod=a*(b对mod 的逆元);求逆元可以用扩欧和费马小定理（可以啊、百度的）；



代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e6+5,mod=1000003;
typedef long long LL;
LL fact[maxn];
void f()
{
    fact[1]=fact[0]=1;
    for(int  i=2;i<maxn;i++)
        fact[i]=(LL)(i*fact[i-1])%mod;
}
LL niyuan(LL a,LL p)//就是个快速幂
{
    if(p==0)return 1;
    LL x=niyuan(a,p/2);
    LL ans=x*x%mod;
    if(p%2==1)ans=ans*a%mod;
    return ans;
}
LL c(int n,int k)
{
    LL fm=(fact[k]*fact[n-k])%mod;
    LL ans1=niyuan(fm,mod-2);//费马小定理，求一个幂就好；
    return (ans1*fact
)%mod;
}
int main()
{
    int t,n,k,kase=0;
    f();//打表
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        if(k*2>n)
            k=n-k;//组合数对称性，减少计算；
        printf("Case %d: %lld\n",++kase,c(n,k));
    }
    return 0;
}