# A类
# 题目链接：
http://codeforces.com/contest/1088/problem/C
# 原题：
You're given an array a of length n. You can perform the following operations on it:
choose an index i (1≤i≤n), an integer x (0≤x≤106), and replace aj with aj+x for all (1≤j≤i), which means add x to all the elements in the prefix ending at i.
choose an index i (1≤i≤n), an integer x (1≤x≤106), and replace aj with aj%x for all (1≤j≤i), which means replace every element in the prefix ending at i with the remainder after dividing it by x.
Can you make the array strictly increasing in no more than n+1 operations?
**输入格式：**
The first line contains an integer n (1≤n≤2000), the number of elements in the array a.
The second line contains n space-separated integers a1, a2, …, an (0≤ai≤105), the elements of the array a.
**输出格式：**
On the first line, print the number of operations you wish to perform. On the next lines, you should print the operations.
To print an adding operation, use the format "1 i x"; to print a modding operation, use the format "2 i x". If i or x don't satisfy the limitations above, or you use more than n+1 operations, you'll get wrong answer verdict.
**输入样例1：**
3
1 2 3
**输出样例1：**
0
**输入样例2：**
3
7 6 3
**输出样例2：**
2
1 1 1
2 2 4
------------
# 题意：
对n个数进行操作，1操作是对1到i加x，2操作是对1到i模x，让你严格输出n+1此操作，让n个数严格递增。
# 思路：
对于n个数化成0，1，2……n-1，的形式，就是满足题意的。
如何化成以上形式呢？
如果一个数不是n的倍数的话，先将它补成n的倍数，然后再加上他的i-1，模n，就成为了i-1.
即先用n次操作变成模n递增，然后直接模n即可。
# AC代码：
```
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
#include<algorithm>
#include<queue>
typedef long long ll;
#include<vector>
#define cin(n) scanf("%lld",&(n))
#define cout(n) printf("%lld",(n))
#define couc(c) printf("%c",(c))
#define coutn printf("\n")
#define cout_ printf(" ")
#define debug() printf("haha\n")
const int MAXN= 1e6 + 5 ;
ll t;
ll n,k;
ll b[MAXN]={1};
ll a[MAXN]={1,1,1,2,3};
ll rou=3;
ll now=5;
ll an;
int main()
{
cin(n);
for(int i=1;i<=n;i++)
cin(a[i]);
ll tmp=0;
cout(n+1);
coutn;
for(int i=n;i>=1;i--)
{
ll x=(i-1)-(a[i]%n)+n;
printf("1 %d %lld\n",i,x);
for(int j=1;j<=i;j++)
a[j]+=x;
}
printf("2 %lld %lld\n",n,n);
return 0;
}
```