51 nod 1189 阶乘分数

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1189
题目思路：
1/n! = 1/x +1/y

==> n! * (x + y) = x * y（右边通分。然后转化）

==> n!^2 = (x - n!)*(y - n!)（左右两边加上n方）

==> a = b * c ,a = n!^2 ,b = x - n! ,c = y - n!

#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
const int maxn=1e6+5;
bool prime[maxn];
int p[maxn/10];
int k;
void isprime()
{
    k=0;
    LL i,j;
    memset(prime, true, sizeof(prime));
    for(i=2; iif(prime[i])
        {
            p[k++]=i;
            for(j=i*i; jfalse;
        }
    }
    /*for(int i=0; i<10; i++)
    cout<
}
LL get(LL m, LL p)
{
    if(mreturn 0;
    return m/p+get(m/p,p);
}
LL quickmod(LL a, LL b)
{
    LL ans = 1;
    while(b)
    {
        if(b&1) ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans ;
}
int main()
{
    isprime();
    int n;
    while(~scanf("%d",&n))
    {
        LL ans=1;
        LL m = quickmod(2,(LL)mod-2);
        for(int i=0; i2+1)%mod;
            ans=ans*tmp%mod;
        }
        ans = ans*m%mod;
        ans = (ans+m%mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}