Codeforces Round #359 (Div. 2) B. Little Robber Girl's Zoo

B. Little Robber Girl's Zoo

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the order of non-decreasing height. However, the animals were so scared that they couldn't stay in the right places.

The robber girl was angry at first, but then she decided to arrange the animals herself. She repeatedly names numbers
l and r such that
r - l + 1 is even. After that animals that occupy positions between
l and r inclusively are rearranged as follows: the animal at position
l swaps places with the animal at position
l + 1, the animal l + 2 swaps with the animal
l + 3, ..., finally, the animal at position
r - 1 swaps with the animal
r.

Help the robber girl to arrange the animals in the order of non-decreasing height. You should name at most
20 000 segments, since otherwise the robber girl will become bored and will start scaring the animals again.

Input
The first line contains a single integer n (1 ≤ n ≤ 100) — number of animals in the robber girl's zoo.

The second line contains n space-separated integers
a1, a2, ..., an (1 ≤ ai ≤ 109),
where ai is the height of the animal occupying the
i-th place.

Output
Print the sequence of operations that will rearrange the animals by non-decreasing height.

The output should contain several lines, i-th of the lines should contain two space-separated integers
li and
ri (1 ≤ li < ri ≤ n) — descriptions of segments
the robber girl should name. The segments should be described in the order the operations are performed.

The number of operations should not exceed 20 000.

If the animals are arranged correctly from the start, you are allowed to output nothing.

Examples

Input

4
2 1 4 3

Output

1 4

Input

7
36 28 57 39 66 69 68

Output

1 46 7

Input

5
1 2 1 2 1

Output

2 5
3 4
1 41 4

Note
Note that you don't have to minimize the number of operations. Any solution that performs at most
20 000 operations is allowed.

题意：给你一个序列，然后你可以选择一个偶数区间[l,r]，使得swap(l,l+1)...swap(r-1,r)，到最后要使这个序列成为一个非递减序列，求经过的步骤，没有最优解限制

思路：一阵乱搞，从头开始遍历，当两个数交换的时候，会跳过，否则一个一个走，直到序列完成，注意特判序列数为1的情况

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int a[MAXN];
int main()
{
int l,r,n;scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
if(n==1)
return 0;
int bz=1;
while(bz){
l=r=1;
for(int i=1;i<n;){
if(a[i]>a[i+1])
swap(a[i],a[i+1]),r=i+1,i+=2;
else{
if((r-l+1)%2==0)
printf("%d %d\n",l,r),l=r=i+1;
else{
if(l==r) l=r=i+1;
else r++;
}
i++;
}
int flag=0;
for(int j=2;j<=n;j++)
if(a[j]<a[j-1]){
flag=1;break;
}
if(flag==0) bz=0;
}
if(l!=r&&(r-l+1)%2==0)
printf("%d %d\n",l,r);
if(bz==0) break;
}
return 0;
}