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[100005];
int b[100005];
bool cmp(node a,node b)
{
    return a.k>b.k;
}

int main()
{
    scanf("%d",&t);
    for(;t>0;t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i].k);
            a[i].pos=i;
        }
        sort(a+1,a+1+n,cmp);
        memset(b,-1,sizeof(b));
        for(int j=2;j<=n;j++)
        {
            if(a[1].pos%j!=0)
                b[j]=a[1].k;
        }
        for(int i=2;i<=n;i++)
        {
            if(b[i]!=-1)
                continue;
            for(int j=1;j<=n;j++)
            if (a[j].pos%i!=0) 
            {
                b[i]=a[j].k;
                break;
            }
        }
        printf("%d",b[2]);
        for(int j=3;j<=n;j++)
            printf(" %d",b[j]);
        printf("
");
    }
    return 0;
}