HDU 6098 17多校6 Inversion（思维+优化）

Problem Description

Give an array A, the index starts from 1.
Now we want to know Bi=maxi∤jAj , i≥2.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integer n : the size of array A.
Next one line contains n integers, separated by space, ith number is Ai.

Limits
T≤20
2≤n≤100000
1≤Ai≤1000000000
∑n≤700000

 

Output

For each test case output one line contains n-1 integers, separated by space, ith number is Bi+1.

 

Sample Input

2

4

1 2 3 4

4

1 4 2 3

 

Sample Output

3 4 3

2 4 4

 

题意：给出a序列，求b序列，bj=max(ai)(i%j!=0)

题解：对a排序后，直接查找TLE。因此将排序后的最大值，将b中可以取它的值先赋值赋好。然后剩余数再去查找，就会节省很多时间。应该有更好的方法？

 

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<queue>
 #include<map>
 #include<vector>
 #include<cmath>
 #include<cstring>

 using namespace std;
 int t,n;
 struct  node
 {
     long long k;
     int pos;
 }a[];
 int b[];
 bool cmp(node a,node b)
 {
     return a.k>b.k;
 }

 int main()
 {
     scanf("%d",&t);
     for(;t>;t--)
     {
         scanf("%d",&n);
         for(int i=;i<=n;i++)
         {
             scanf("%lld",&a[i].k);
             a[i].pos=i;
         }
         sort(a+,a++n,cmp);
         memset(b,-,sizeof(b));
         for(int j=;j<=n;j++)
         {
             if(a[].pos%j!=)
                 b[j]=a[].k;
         }
         for(int i=;i<=n;i++)
         {
             if(b[i]!=-)
                 continue;
             for(int j=;j<=n;j++)
             if (a[j].pos%i!=)
             {
                 b[i]=a[j].k;
                 break;
             }
         }
         printf("%d",b[]);
         for(int j=;j<=n;j++)
             printf(" %d",b[j]);
         printf("\n");
     }
     return ;
 }