Ehab and a 2-operation task【数论思想】
You're given an array aa of length nn. You can perform the following operations on it:
- choose an index ii (1≤i≤n)(1≤i≤n), an integer xx (0≤x≤106)(0≤x≤106), and replace ajaj with aj+xaj+x for all (1≤j≤i)(1≤j≤i), which means add xx to all the elements in the prefix ending at ii.
- choose an index ii (1≤i≤n)(1≤i≤n), an integer xx (1≤x≤106)(1≤x≤106), and replace ajaj with aj%xaj%x for all (1≤j≤i)(1≤j≤i), which means replace every element in the prefix ending at ii with the remainder after dividing it by xx.
Can you make the array strictly increasing in no more than n+1n+1 operations?
Input
The first line contains an integer nn (1≤n≤2000)(1≤n≤2000), the number of elements in the array aa.
The second line contains nn space-separated integers a1a1, a2a2, ……, anan (0≤ai≤105)(0≤ai≤105), the elements of the array aa.
Output
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 "11 ii xx"; to print a modding operation, use the format "22 ii xx". If ii or xx don't satisfy the limitations above, or you use more than n+1n+1 operations, you'll get wrong answer verdict.
Examples
Input
3
1 2 3
Output
0
Input
3
7 6 3
Output
2
1 1 1
2 2 4
Note
In the first sample：
the array is already increasing so we don't need any operations.
In the second sample:
In the first step: the array becomes [8,6,3][8,6,3].
In the second step: the array becomes [0,2,3][0,2,3].
思路：
先将输入的 a[i] 全部加上 MAX=1e6
然后从i=1 开始到 i=n 每次将 a[i]%(a[i]-i)
这样可以保证每次取余前面的每个值都不受影响（怎么也想不到这样……）
例如：
a[i]： 3 6 4 9
a[i]+MAX： 1000003 1000006 1000004 1000009
a[i]%(a[i]-i) ： 1 2 3 4
AC代码：
#include<stdio.h>
const int MAX=1e6;
int main()
{
int a[MAX+5],n;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
printf("%d\n",n+1);
printf("1 %d %d\n",n,MAX);
for(int i=0;i<n;i++){
printf("2 %d %d\n",i+1,a[i]+MAX-(i+1));
}
return 0;
}
