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 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
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
[cpp]
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#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; }
#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;
}
