Decrease (Contestant ver.) AtCoder - 2661

标签：iostream   var   nbsp   opera   const   val   value   lse   最小   

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

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.

Sample Input 1

0

Sample Output 1

4
3 3 3 3

Sample Input 2

1

Sample Output 2

3
1 0 3

Sample Input 3

2

Sample Output 3

2
2 2

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

Sample Input 4

3

Sample Output 4

7
27 0 0 0 0 0 0

Sample Input 5

1234567894848

Sample Output 5

10
1000 193 256 777 0 1 1192 1234567891011 48 425
嗯，找规律，可惜找不不到，然后经讲解恍然大悟，只需要最小值加n 剩下的减一就好了（就是递归 k-1成立k就成立）
于是：

#include <iostream>
using namespace std;
int main()
{
    long long c[51];
    long long k,a,b;
    int j=0;
    scanf("%lld",&k);
    a=k/50;
    b=k%50;
    cout<<"50"<<endl;
    if(k%50==0)
	{
        for(long long i=50+a-1;i>=a;i--)
            cout<<i<<" ";
    }
    else
	{
        for(long long i=50+a;i>=a;i--)
		{
            c[j]=i;
            j++;
        }
        for(long long i=0;i<51;i++)
		{
            if(i==b)
                continue;
            else
                cout<<c[i]<<" ";
        }
    }
    cout<<endl;
    return 0;
}
/*
    49 48 47.........0    K=0
50     48 47.........0    K=1
50  49    47.........0    K=2
    50 49 48.........1    K=50
*/

Decrease (Contestant ver.) AtCoder - 2661

标签：iostream   var   nbsp   opera   const   val   value   lse   最小   

原文地址：https://www.cnblogs.com/mozheaishang/p/10017449.html