hdu5407CRB and Candies (逆元+数学公式)

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
   
   
   
   

题意：求LCM(C(n,0),C(n,1),C(n,2),...,C(n,n)) (LCM指的是最小公倍数)
思路：一开始想每次求两个数的最小公倍数，然后求得n个数的最小公倍数，结果发现打表打不出= =。看了别人思路，发现求的式子是一个数学公式
令a[n]=LCM(C(n,0),C(n,1),C(n,2),...,C(n,n))
b[n]=LCM(1,2,3,...,n)
a[n]=b[n+1]/(n+1)
if(n=p^k) bn=p*bn-1 else bn=bn-2 p为素数，符合要求的n如4,8,9,25
所以我们可以先把素数筛选出来，并判断1~n这些数是不是等于p^k,把a[]数组预处理出来，然后用逆元就行了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 1000050
#define MOD 1000000007
int prime[maxn];
ll inv[maxn];
ll a[maxn];
void shake(){
    int i;
    inv[1]=1;
    for(i=2;i<=1000000;i++){
        inv[i]=(MOD-MOD/i)*inv[MOD%i]%MOD;
    }
}
ll gcd(ll a,ll b){
    return b ? gcd(b,a%b) : a;
}
int ok(int x)
{
    int t=x,i,j;
    while(t){
        if(t%prime[x]==0)t/=prime[x];
        else break;
    }
    if(t==1)return 1;
    else return 0;
}
void init()
{
    int i,j;
    for(i=1;i<=1000000;i++)prime[i]=i;
    for(i=2;i<=1000000;i++){
        if(prime[i]==i){
            for(j=i+i;j<=1000000;j+=i){
                prime[j]=i;
            }
        }
    }
    a[1]=1;
    for(i=2;i<=1000000;i++){
        if(ok(i)){
            a[i]=a[i-1]*prime[i]%MOD;
        }
        else a[i]=a[i-1];
    }
}
int main()
{
    int T,i,j;
    ll n,m,ans,num;
    shake();
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&n);
        printf("%lld\n",a[n+1]*inv[n+1]%MOD );
    }
    return 0;
}