题目链接
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?
Input
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.
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 “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.
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].
In the second step: the array becomes [0,2,3].
题解:我把数组中的每个数都加上一个1e5 然后再对其中的每一个数进行取余，变成从 0 1 … n - 1 的一个序列
#include<bits/stdc++.h>
using namespace std;
const int N = 2005;
int n,a[N];
int main(){
scanf("%d",&n);for(int i = 1;i <= n;i++) scanf("%d",a + i);printf("%d\n1 %d 100000\n",n + 1,n);for(int i = 1;i <= n;i++)printf("2 %d %d\n",i,a[i] + 100000 - i);return 0;
}