HDU - 6098：Inversion（暴力均摊）

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

, i≥2 i≥2

.

InputThe 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 A i  Ai

.

Limits
T≤20 T≤20

2≤n≤100000 2≤n≤100000

1≤Ai≤1000000000 1≤Ai≤1000000000

∑n≤700000 ∑n≤700000

OutputFor each test case output one line contains n-1 integers, separated by space, ith number is B i+1  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：对于所有的i（i!=1），找最大的a[j]，满足j%i!=0；

思路：没有什么对应的算法，居然是暴力，我们从大到小排序，然后找到第一个满足题意的即可。

复杂度均摊下来是线性的，显然过得去。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
struct in{
    int id,num;
    friend bool operator <(in w,in v) {return w.num>v.num;}
}s[];
int main()
{
    int T,N;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        rep(i,,N) scanf("%d",&s[i].num),s[i].id=i;
        sort(s+,s+N+);
        rep(i,,N) {
            rep(j,,N){
                if(s[j].id%i!=){
                    printf("%d ",s[j].num); break;
                }
            }
        }
        puts("");
    }
    return ;
}