/*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