D - Decrease (Contestant ver.) AtCoder - 2661

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N?1 or smaller.

• Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N?1 or smaller after a finite number of operations.

You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.

• 0≤K≤50×10¹⁶

Input is given from Standard Input in the following format:

K

Print a solution in the following format:

N
a1 a2 ... aN

Here, 2≤N≤50 and 0≤a_i≤10¹⁶+1000 must hold.

0

4
3 3 3 3

1

3
1 0 3

2

2
2 2

The operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].

3

7
27 0 0 0 0 0 0

1234567894848

10
1000 193 256 777 0 1 1192 1234567891011 48 425
思路：找规律发现
可以用50个数存，逆推，初始数组为：【0-49】
每经过50次操作，数组的每个数都会+1
然后要计算的就是k%50的操作
找规律发现，i<余数时，每个数+51-余数
i>余数时，每个数-余数
代码如下：

 1 #include<bits/stdc++.h>
 2 #include<iostream>
 3 using namespace std;
 4 #define ll long long
 5 
 6 int main()
 7 {
 8     ll k;
 9     while(cin >> k)
10     {
11         ll ro = k / 50;
12         ll yu = k % 50;
13         ll ty = ro - yu;
14         cout << 50 << endl;
15         for(int i = 0;i < 50;i++)
16         {
17             if(i < yu)
18             cout <<  i + ty + 51;
19             else
20             cout << i + ty;
21             if(i != 49)
22             cout << " ";
23         }
24         cout << endl;
25     }
26     return 0;
27 }
28 /*
29 1
30 50
31 50 0 1 2 3 4 5 6 7 8
32 9 10 11 12 13 14 15 16 17 18
33 19 20 21 22 23 24 25 26 27 28
34 29 30 31 32 33 34 35 36 37 38
35 39 40 41 42 43 44 45 46 47 48
36 */