数论 HDOJ 5407 CRB and Candies

题目传送门

题意：求LCM (C(N,0),C(N,1),...,C(N,N))，LCM是最小公倍数的意思，C函数是组合数。

分析：先上出题人的解题报告

　　

　　好吧，数论一点都不懂，只明白f (n + 1)意思是前n+1个数的最小公倍数，求法解释参考HDOJ 1019，2028

这个结论暂时不知道怎么推出来的，那么就是剩下1/(n+1) 逆元的求法了

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-21 14:52:39
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
ll f[N];
int p[N];

bool ok(int n)  {
    int t = p[n];
    while (n % t == 0 && n > 1) n /= t;
    return n == 1;
}

void seive(int n)   {
    for (int i=1; i<=n; ++i)    p[i] = i;
    for (int i=2; i<=n; ++i)    {
        if (p[i] == i)    {
            for (int j=2*i; j<=n; j+=i) p[j] = i;
        }
    }
}

ll pow_mod(int a, int x, int p) {
    ll ret = 1;
    while (x)   {
        if (x & 1)  ret = ret * a % p;
        a = a * 1ll * a  % p;
        x >>= 1;
    }
    return ret;
}

ll Inv(int a)   {
    return pow_mod (a, MOD - 2, MOD);
}

void solve(void)    {
    seive (1000010);
    f[0] = 1;
    for (int i=1; i<=1000010; ++i)  {
        if (ok (i))    {
            f[i] = f[i-1] * p[i] % MOD;
        }
        else    f[i] = f[i-1];
    }
}

int main(void)    {
    solve ();
    int T;  scanf ("%d", &T);
    while (T--) {
        int n;  scanf ("%d", &n);
        printf ("%I64d\n", f[n+1] * Inv (n + 1) % MOD);
    }

    return 0;
}