HDU 6098 Inversion

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

Source
2017 Multi-University Training Contest - Team 6

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b[i]表示a數組中下標不能被i整除的最大值。
經過排序之後，之後那麼每次找到的第一個數就是答案了，時間複雜度低於O(n*n)

#include<iostream>#include<algorithm>using namespace std;//不要小瞧人生啊const int N = 1e5 + 10;int T, n, a[N], id[N];int main() {    scanf("%d", &T);    for(int i=0;i<T;i++)    {        scanf("%d", &n);        for(int i=1;i<n+1;i++)            scanf("%d", a + i), id[i] = i;        sort(id + 1, id + 1 + n, [&](int x, int y)         {            return a[x]>a[y];         });//又學習到了cmp的另一種套路 //[&](int x,int y){return a[x]>a[y]}        for(int i=2;i<n+1;i++)//暴力        {            int f = 0;            for(int j=1;j<n+1;j++)                if (id[j] % i != 0) //下標                {                f = a[id[j]];                break;                }            printf("%d%c", f, " \n"[i == n]);        }    }    return 0;}