hdu6098
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Inversion
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 271 Accepted Submission(s): 183
Problem Description
Give an array A, the index starts from 1.
Now we want to knowBi=maxi∤jAj , i≥2 .
Now we want to know
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 isAi .
Limits
T≤20
2≤n≤100000
1≤Ai≤1000000000
∑n≤700000
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
Limits
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
——————————————————————————————————
题目的意思是给出一个序列,求2~n每一个数,下标不是这个数倍数的最大值是什么?
思路:从大到小排个序,然后枚举判下表是否为这个数的倍数
- #include <iostream>
- #include <cstdio>
- #include <cstring>
- #include <map>
- #include <set>
- #include <string>
- #include <cmath>
- #include <algorithm>
- #include <vector>
- #include <bitset>
- #include <stack>
- #include <queue>
- #include <unordered_map>
- #include <functional>
- using namespace std;
- struct node{
- int id,v;
- }a[100005];
- bool cmp(node x,node y)
- {
- return x.v>y.v;
- }
- int main()
- {
- int T,n;
- for(scanf("%d",&T);T--;)
- {
- scanf("%d",&n);
- for(int i=1;i<=n;i++)
- scanf("%d",&a[i].v),a[i].id=i;
- sort(a+1,a+1+n,cmp);
- int q=0;
- for(int i=2;i<=n;i++)
- {
- int k=1;
- while(a[k].id%i==0)
- {
- k++;
- }
- if(q++)
- printf(" ");
- printf("%d",a[k].v);
- }
- printf("\n");
- }
- return 0;
- }
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