/*1718ms,90976KB*/
#include<cstdio>
#include<cmath>
const long long mod = 1000000007;
const int mx = 10000001;
const int sqrtmx = (int)sqrt((double)mx);
long long fac[mx] = {1, 1};
bool vis[mx];
int prime[664580], cnt;
void init()
{
for (int i = 2; i < mx; ++i)
{
fac[i] = fac[i - 1];
if (!vis[i])
{
prime[cnt++] = i;
if (i <= sqrtmx)
for (int j = i * i; j < mx; j += i)
vis[j] = true;
}
else fac[i] = (fac[i] * i) % mod;
}
}
int main()
{
init();
int n;
while (scanf("%d", &n), n)
{
long long res = fac[n];
for (int i = 0; i < cnt && prime[i] <= (n >> 1); ++i)
{
int r = 0, tmp = n;
while (tmp) r += (tmp /= prime[i]);
if ((r & 1) == 0) res = (res * prime[i]) % mod;
}
printf("%lld\n", res);
}
return 0;
}
SWERC 2011 / HDU 4196 Remoteland (数论&想法题)
原文:http://blog.csdn.net/synapse7/article/details/18231709