hdu 多校联赛 Inversion

Inversion

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

241 2 3 441 4 2 3

 

Sample Output

3 4 32 4 4

 

题意比较简单 在这也就不再重复 这道题出错大多是tle 其实题目已经告诉了本题的解法 反转 正着推超时 那就反过来推

ac代码：

#include<cstdio>#include<iostream>#include<cstring>#include<stdlib.h>using namespace std;const int N=1e5+10;struct Node{    int val;    int pos;}A[N];int cmpDec(const void *a,const void *b)//怕超时特地用了快排 其实sort也能过{    return (((Node*)b)->val)-(((Node*)a)->val);}int main(){    int t;    cin>>t;    int n;    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            {             scanf("%d",&A[i].val);             A[i].pos=i;            }        qsort(A+1,N,sizeof(Node),cmpDec);        for(int i=2;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(A[j].pos%i!=0)                {                    if(i==2)                    {                        printf("%d",A[j].val);                        break;                    }                    else                    {                        printf(" %d",A[j].val);                        break;                    }                }            }        }        printf("\n");    }    return 0;}