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Hdu4630: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4630

Question: calculate the maximum GCD range for a given sorting.

Question: discrete tree array. First, sort the query by the Left endpoint from large to small. Then, the tree array is used to maintain the maximum number of common approx. Enumerative the approximate number of each number. record the position B [x] At a multiple of the previous x until the current position.

 1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<vector> 6 using namespace std; 7 const int N=5e4+19; 8 int a[N],c[N],b[N],ans[N]; 9 int n,m;10 int lowbit(int x){11    return x&(-x);12 }13 void add(int x,int val){14     while(x<=n){15         c[x]=max(c[x],val);16         x+=lowbit(x);17     }18 }19 int getsum(int x){20    int sum=0;21    while(x>0){22      sum=max(sum,c[x]);23      x-=lowbit(x);24    }25    return sum;26 }27 struct Node{28   int l,r;29   int id;30   bool operator<(const Node a)const{31     return l>a.l;32   }33 }num[N];34 int main(){35     int T;36     scanf("%d",&T);37     while(T--){38         scanf("%d",&n);39         memset(a,0,sizeof(a));40         memset(c,0,sizeof(c));41         memset(b,0,sizeof(b));42         for(int i=1;i<=n;i++)43           scanf("%d",&a[i]);44           scanf("%d",&m);45           for(int i=1;i<=m;i++){46             scanf("%d%d",&num[i].l,&num[i].r);47             num[i].id=i;48           }49         sort(num+1,num+m+1);50         int j=n;51         for(int i=1;i<=m;i++){52             for(;j>=num[i].l;j--){53                for(int k=1;k*k<=a[j];k++){54                    if(a[j]%k==0){55                        if(b[k]){56                         add(b[k],k);57                        }58                         b[k]=j;59                       if(k*k!=a[j]){60                          if(b[a[j]/k]){61                             add(b[a[j]/k],a[j]/k);62                          }63                          b[a[j]/k]=j;64                       }65                    }66                }67              }68            ans[num[i].id]=getsum(num[i].r);69          }70      for(int i=1;i<=m;i++)71         printf("%d\n",ans[i]);72     }73 }

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No pain no game