[數學]hdu5407 CRB and Candies

CRB and Candies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 209    Accepted Submission(s): 84

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

1002.CRB and Candies

The problem is just to calculate $g(N)=LCM(C(N,0),C(N,1),...,C(N,N))$.

Introducing function $f(n)=LCM(1,2,...,n)$, the fact $g(n)=f(n+1)/(n+1)$holds.

We calculate $f(n)$ in the following way.

$f(1)=1$.

If n\ =p^{k}n =p​k​​ then $f(n)=f(n-1)\times p$, else $f(n)=f(n-1)$.

Time complexity:$O(N\cdot logN)$

#include <cstdio>
#include <iostream>
#include <algorithm>
#include<cstring>
using namespace std;

int n,T;
const int MAXN = 1000100;
typedef long long ll;
ll f[MAXN],p[MAXN];
const int mod = 1000000007;
ll pmod(ll a, ll n)
{
    ll ret = 1;
    for (; n; n>>=1, a=a*a%mod) if (n & 1)
        ret = ret * a % mod;
    return ret;
}
bool ok(ll n)
{
    int t = p[n];
    while (n % t==0 && n > 1) n /= t;
    return n==1;
}
ll inv(ll a)
{
    return pmod(a, mod - 2);
}
void init()
{
    for(int i = 1;i<MAXN;i++) p[i] = i;
    for(int i = 2;i<MAXN;i++)
        if (p[i] == i)
        {
            for(int j = i+i; j <MAXN;j+=i)
                p[j] = i;
        }
    f[0] = 1;
    for(int i = 1;i<MAXN;i++)
    {
        if(ok(i))
            f[i] = f[i-1] * p[i] % mod;
        else
            f[i] = f[i-1];
    }
//    for(int i = 1;i<=200;i++)
//        printf("%d %d\n",i,p[i]);
}

int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        n++;
        ll ans =  f[n] * inv(n) % mod;
        printf("%I64d\n",ans);
    }
    return 0;
}