Codeforces 359D Pair of Numbers | 二分+ST表+gcd

题面:

给一个序列,求最长的合法区间,合法被定义为这个序列的gcd=区间最小值

输出最长合法区间个数,r-l长度

接下来输出每个合法区间的左端点

---

题解:

由于区间gcd满足单调性,所以我们可以二分区间长度,用st表维护区间最小值和gcd即可

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #define N 300100
 using namespace std;
 int gcd(int x,int y)
 {
     return y==?x:gcd(y,x%y);
 }
 int rgcd[N][],rmin[N][],n,l,r,mid,ans,ok[N];
 int check(int len)
 {
     if (len==) return ;
     for (int i=;i+len<=n;i++)
     {
     int tmp=log2(len+);
     if (gcd(rgcd[i][tmp],rgcd[i+len-(<<tmp)+][tmp]) == min(rmin[i][tmp],rmin[i+len-(<<tmp)+][tmp]))
         return ;
     }
     return ;
 }
 int main()
 {
     scanf("%d",&n);
     for (int i=,x;i<=n;i++)
     scanf("%d",&x),rgcd[i][]=rmin[i][]=x;
     for (int j=;j<=;j++)
     for (int i=;i+(<<j)-<=n;i++)
         rgcd[i][j]=gcd(rgcd[i][j-],rgcd[i+(<<j-)][j-]),rmin[i][j]=min(rmin[i][j-],rmin[i+(<<j-)][j-]);
     r=*n;
     while (l<r)
     {
     mid=l+r+>>;
     if (check(mid))
         l=mid;
     else r=mid-;
     }
     for (int i=;i+l<=n;i++)
     {
     int tmp=log2(l+);
     if (gcd(rgcd[i][tmp],rgcd[i+l-(<<tmp)+][tmp]) == min(rmin[i][tmp],rmin[i+l-(<<tmp)+][tmp]))
         ok[i]=,ans++;
     }
     printf("%d %d\n",ans,l);
     for (int i=;i<=n;i++)
     if (ok[i]) printf("%d ",i);
     return ;
 }