Atcoder At Beginner Contest 068 D - Decrease (Contestant ver.)

D - Decrease (Contestant ver.)

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

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.

Constraints

• 0≤K≤50×10¹⁶

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Input

Input is given from Standard Input in the following format:

K

Output

Print a solution in the following format:

N
a1 a2 ... aN

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

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Sample Input 1

Copy

0

Sample Output 1

Copy

4
3 3 3 3

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Sample Input 2

Copy

1

Sample Output 2

Copy

3
1 0 3

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Sample Input 3

Copy

2

Sample Output 3

Copy

2
2 2

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

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Sample Input 4

Copy

3

Sample Output 4

Copy

7
27 0 0 0 0 0 0

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Sample Input 5

Copy

1234567894848

Sample Output 5

Copy

10
1000 193 256 777 0 1 1192 1234567891011 48 425

直接定义数组长为50,k=0:0,1,2,............49;
k=1:50,0,1,2,3,..............48;
k=2:49,50,1,2,3..............47;
.......
k=50:1,2,3,..............48,49,50;
周期为50

#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath> 
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f 
#define mem(a) (memset(a,0,sizeof(a))) 
typedef long long ll;
ll n;
ll a[55];
int main()
{
    scanf("%lld",&n);
    for(int i=1;i<=50;i++)
    {
        a[i]=n/50+i-1;
    }
    n%=50;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=50;j++)
        {
            a[j]--;
        }
        a[i]+=51;
    }
    printf("50\n");
    for(int i=1;i<=50;i++)
    {
        if(i!=1) printf(" ");
        printf("%lld",a[i]);
    }
    printf("\n");
    return 0;
}