hdu6098Inversion

Inversion

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 602    Accepted Submission(s): 401

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

主要是看懂公式： maxi∤jAj

 

i不能整除j

由此有兩種方式解決，可以用結構體排序，將權值大的排在前面，再對下標進行篩選

還可以直接枚舉，但有一個優化，先對a選出最大的數和座標，然後判斷maxa是否滿足，不滿足再從1~n中枚舉篩選

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
#define N 100010
#define LL long long
LL a[N];
LL b[N];
int main()
{
    int t;
    LL n;

    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        scanf("%lld",&n);
        LL maxa=0;
        LL maxa_i=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&a[i]);
            if(a[i]>maxa)
            {
                maxa=a[i];
                maxa_i=i;
            }
        }
        //int maxx;
        for(int i=2;i<=n+1;i++)
        {
            if(maxa_i%i==0)
            {
                LL mm=0;
                for(int j=1;j<=n;j++)
                {
                    if(j%i==0)continue;
                    if(a[j]>mm)
                    {
                        mm=a[j];
                    }
                }
                b[i]=mm;
            }
            else
            b[i]=maxa;
        }
        for(int i=2;i<n;i++)
        {
            cout<<b[i]<<' ';
        }
        cout<<b[n]<<endl;
    }
    return 0;
}

 

maxi∤jAj , i≥2.