CF 683 div.2 A Add Candies
题目：

There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.

To achieve this, you will:

Choose m such that 1≤m≤1000
Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.

Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.

It can be proved that for the given constraints such a sequence always exists.

You don’t have to minimize m.

If there are several valid sequences, you can output any.

Input
Each test contains multiple test cases.

The first line contains the number of test cases t (1≤t≤100). Description of the test cases follows.

The first and only line of each test case contains one integer n (2≤n≤100).

Output
For each testcase, print two lines with your answer.

In the first line print m (1≤m≤1000) — the number of operations you want to take.

In the second line print m positive integers a1,a2,…,am (1≤ai≤n), where aj is the number of bag you chose on the j-th operation.

Example
inputCopy
2
2
3
outputCopy
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2,2] candies.

In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7,8,3]. Later, you add 4 candies to second and third bag, so you have [7,12,7], and 5 candies to first and third bag — and the result is [12,12,12].
题解：
思维题，自己写一下就好。

#include <stdio.h>
int main()
{
  int t;
  scanf("%d",&t);
  while(t--){
    int n;
    scanf("%d",&n);
    printf("%d\n",n-1);
    for(int i=2;i<=n;i++){
      printf("%d",i);
      if(i!=n) printf(" ");
      else printf("\n");
    }
  }
  return 0;
}