[思路题] Codeforces 354C #206 (Div. 1) C. Vasya and Beautiful Arrays

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枚举gcd为g 然后再枚举g的所有倍数g∗i 统计下在 [g∗i,g∗i+K]范围内共有几个数总数为n即合法

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=1000005;int n,K;int a[N],c[N];int main(){  int m=0;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(K);  for (int i=1;i<=n;i++) read(a[i]),m=max(m,a[i]),c[a[i]]++;  for (int i=1;i<=m;i++) c[i]+=c[i-1];  for (int i=m;i>1;i--){    int sum=0;    for (int j=i;j<=m;j+=i)      sum+=c[min(m,min(j+i-1,j+K))]-c[j-1];    if (sum==n)      return printf("%d\n",i),0;  }  printf("1\n");  return 0;}

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