HDU 6098 Inversion

Inversion

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 963    Accepted Submission(s): 614

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

 

Source

2017 Multi-University Training Contest - Team 6

 

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找个结构体，里面存上数值和下标，然后俺数值排序，从最大值开始找就可以了

My ugly code

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <iostream>#include <algorithm>#define ll long longusing namespace std;int T,N;int a,b,c,d,e,f,g;int main(){    scanf("%d",&T);    while(T--){        scanf("%d",&N);        int ans[10];        int kk=0;        for(int i=1;i<=N;i++){            scanf("%d%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f,&g);            ans[0]=d-g;            ans[1]=e-g;            ans[2]=f-g;            ans[3]=a-ans[0]-ans[2]-g;            ans[4]=b-ans[0]-ans[1]-g;            ans[5]=c-ans[1]-ans[2]-g;            ans[6]=g;            if(ans[0] < 0 || ans[1] < 0 || ans[2] < 0 || ans[3] < 0 || ans[4] < 0 || ans[5] < 0 || ans[6] < 0)                continue ;            int tmp=0;            for(int i=0;i<7;i++)                tmp+=ans[i];            kk=max(kk,tmp);        }        printf("%d\n",kk);    }    return 0;}