[数论] 51Nod 1189 阶乘分数

来源：互联网发布：软件开发前期准备工作编辑：程序博客网时间：2024/04/30 10:50

(X-N!)(Y-N!) = (N!)^2

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }  return *p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=1000005;int prime[N],vst[N],num;inline void Pre(int maxn){  for (int i=2;i<=maxn;i++){    if (!vst[i]) prime[++num]=i;    for (int j=1;j<=num && (ll)prime[j]*i<=maxn;j++){      vst[prime[j]*i]=1;      if (i%prime[j]==0)break;    }  }}int n;int q[N];ll ans=1;const int P=1e9+7;const int INV=5e8+4;int main(){  scanf("%d",&n); Pre(n);  for (int i=1;i<=num;i++){    int tem=prime[i];    q[i]=n/tem;    while ((ll)tem*prime[i]<=n)      tem*=prime[i],q[i]+=n/tem;    (ans*=(2*q[i]+1))%=P;  }  printf("%lld\n",((ans+P-1)%P*INV%P+1)%P);  return 0;}

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