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

2
4
1 2 3 4
4
1 4 2 3

Sample Output

3 4 3
2 4 4

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题意：

数组Bi的值等于下标i不整除j的最大的数组A中的值，直接给个样例解释

以1 4 2 3为例 2不能被1 、3整除，A1=1<A3=2，所以B2的值为2；

思路：

用pair将下标与value的值关联起来，以value的值从大到小排序，进行整除比较，找到最大的那个下标不能被整除的值

可以使用多中数据结构，或map或vector或list等

pair的排序重定义函数如下：

bool judge(const pair<int,int> a,const pair<int,int> b)

{

return a.first>b.first;//从大到小排序

}

代码如下：

#include<iostream>

#include<list>

#include<vector>

#include<string.h>

#include<algorithm>

#include<stdio.h>

using namespace std;

const int maxn=100006;

typedef long long ll;

ll b[maxn];

bool judge(const pair<ll,ll> a,const pair<ll,ll>b)

{

    return a.first>b.first;

}

int main()

{

    int t;

   scanf("%d",&t);

    ll a;

    int n;

    while(t--)

    {

        vector<pair<ll,ll> >vec;

        memset(b,0,sizeof(b));

        scanf("%d",&n);

        for(ll i=1;i<=n;i++)

        {

           scanf("%lld",&a);

            vec.push_back(make_pair(a,i));

        }

        sort(vec.begin(),vec.end(),judge);

        for(ll j=2;j<=n;j++)

        {

            for(ll i=0;i<vec.size();i++)

            {

                if(vec[i].second%j!=0)

                {

                    b[j]=vec[i].first;

                    break;

                }

                else continue;

            }

        }

        for(ll i=2;i<n;i++)

            printf("%lld ",b[i]);

       printf("%lld\n",b[n]);

    }

    return 0;

}

BI

2017杭电多校第六场03Inversion

Inversion