hdu 5407【LCM性质】+【逆元】（结论题）

<题目链接>

<转载于 >>> >

Problem Description

CRB has  N different candies. He is going to eat K candies.
He wonders how many combinations he can select.
Can you answer his question for all K(0 ≤ K ≤ N)?
CRB is too hungry to check all of your answers one by one, so he only asks least common multiple(LCM) of all answers.

 

Input

There are multiple test cases. The first line of input contains an integer  T, indicating the number of test cases. For each test case there is one line containing a single integer N.
1 ≤ T ≤ 300
1 ≤ N ≤ 106

 

Output

For each test case, output a single integer – LCM modulo 1000000007( 109+7).

 

Sample Input

5

1

2

3

4

5

 

Sample Output

1

2

3

12

10

 

题目大意：

题目大意就是求 ：    lcm(C(n,0),C(n,1),C(n,2),,,,C(n,n))

解题分析：

有一个对应的结论： lcm(C(n,0),C(n,1),C(n,2),,,,C(n,n))  =   lcm(1,2,,,,n,n+1)/(n+1)。

于是这道题就变成了求（1~n）的lcm，当然，直接暴力求解会超时，还有求LCM的更加高效的解法，叫做分解质因数法。并且，由于（n+1）可能很大，所以还要用到逆元的知识。

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long LL;
const int N = 1000005;
const LL mod = 1000000007;
LL f[N];
LL gcd(LL a,LL b){
    return b==0?a:gcd(b,a%b);
}

LL extend_gcd(LL a,LL b,LL &x,LL &y){
    if(!b){
        x=1,y = 0;
        return a;
    }else{
        LL x1,y1;
        LL d = extend_gcd(b,a%b,x1,y1);
        x = y1;
        y = x1 - a/b*y1;
        return d;
    }
}
LL mod_reverse(LL a,LL n)
{
    LL x,y;
    LL d=extend_gcd(a,n,x,y);
    if(d==1) return (x%n+n)%n;
    else return -1;
}
int prime[N];
LL F[N];
bool only_divide(int n){
    int t = prime[n];
    while(n%t==0){
        n/=t;
    }
    if(n==1) return true;
    return false;
}
void init(){
    for(int i=1;i<N;i++){
        prime[i] = i;
    }
    for(int i=2;i<N;i++){  ///十分巧妙的一步,判断某个数是否只有唯一的质因子,只需要把每个数的倍数存下来
        if(prime[i]==i){
            for(int j=i+i;j<N;j+=i){
                prime[j] = i;
            }
        }
    }
    F[1] = 1;
    for(int i=2;i<N;i++){
        if(only_divide(i)){
            F[i] = F[i-1]*prime[i]%mod;
        }else F[i] = F[i-1];
    }
}
int main()
{
    init();
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        int n;
        scanf("%d",&n);
        LL inv = mod_reverse((n+1),mod);
        printf("%lld\n",F[n+1]*inv%mod);
    }
    return 0;
}

2018-07-30