D. Equalize Them All
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array a consisting of n integers. You can perform the following operations arbitrary number of times (possibly, zero):
Choose a pair of indices (i,j) such that |i−j|=1 (indices i and j are adjacent) and set ai:=ai+|ai−aj|;
Choose a pair of indices (i,j) such that |i−j|=1 (indices i and j are adjacent) and set ai:=ai−|ai−aj|.
The value |x| means the absolute value of x. For example, |4|=4, |−3|=3.

Your task is to find the minimum number of operations required to obtain the array of equal elements and print the order of operations to do it.

It is guaranteed that you always can obtain the array of equal elements using such operations.

Note that after each operation each element of the current array should not exceed 1018 by absolute value.

Input
The first line of the input contains one integer n (1≤n≤2⋅105) — the number of elements in a.

The second line of the input contains n integers a1,a2,…,an (0≤ai≤2⋅105), where ai is the i-th element of a.

Output
In the first line print one integer k — the minimum number of operations required to obtain the array of equal elements.

In the next k lines print operations itself. The p-th operation should be printed as a triple of integers (tp,ip,jp), where tp is either 1 or 2 (1 means that you perform the operation of the first type, and 2 means that you perform the operation of the second type), and ip and jp are indices of adjacent elements of the array such that 1≤ip,jp≤n, |ip−jp|=1. See the examples for better understanding.

Note that after each operation each element of the current array should not exceed 1018 by absolute value.

If there are many possible answers, you can print any.

## Examples input 5 2 4 6 6 6 output 2 1 2 3 1 1 2 input 3 2 8 10 output 2 2 2 1 2 3 2 input 4 1 1 1 1 output 0

``````#include<bits/stdc++.h>
using namespace std;

int main()
{
int n,ma=-1,p,f;
int a[200010],b[200010];
memset(b,0,sizeof(b));
cin>>n;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[a[i]]++;
if(b[a[i]]>ma)
{
ma=b[a[i]];
f=a[i];
p=i;
}
}
int t=0;
for(int i=1;i<=n;i++)
if(a[i]!=f) t++;

printf("%d\n",t);

for(int i=p;i>=1;i--)
{
if(a[i]==f)
continue;
else if(a[i]>f)
printf("2 %d %d\n",i,i+1);

else printf("1 %d %d\n",i,i+1);
}
for(int i=p;i<=n;i++)
{
if(a[i]==f)
continue;
else if(a[i]>f) printf("2 %d %d\n",i,i-1);
else printf("1 %d %d\n",i,i-1);
}

}

``````