AtCoder Beginner Contest 068 D

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 ai 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:

Na1 a2 ... aN

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

---

Sample Input 1

0

Sample Output 1

43 3 3 3

---

Sample Input 2

1

Sample Output 2

31 0 3

---

Sample Input 3

2

Sample Output 3

22 2

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

---

Sample Input 4

3

Sample Output 4

727 0 0 0 0 0 0

---

Sample Input 5

1234567894848

Sample Output 5

101000 193 256 777 0 1 1192 1234567891011 48 425

#include <cstdio>#include <cstring>#include <iostream>using namespace std;long long int a[51];int main(){    long long int n;    while(scanf("%lld",&n)==1)    {        memset(a,0,sizeof(a));        for(long long int i=1;i<=50;i++)            a[i]=49+n/50;                      //n/50就相当于50个数字每个数字都得要被减（n/50）次50还要加（n/50）次49（因为每个数被减去其他的数就要加一）        long long int m=n%50;                  //多出来的要执行的操作次数；        long long int j;        for(int i=1;i<=50;i++) a[i]=a[i]-m;    //根据逆向思维，多出多少次，其他的数字就要减去多少个一；        for(j=1;j<=m;j++)a[j]+=51;             //因为多出来m次，所以在m个数上加50；        printf("50\n");        for(int i=1;i<=50;i++)printf("%lld ",a[i]);        printf("\n");    }    return 0;}
//明明不是一道困难的题，之前却一直wa，我之前是把所有的m*50全部加在最后一个数字上可是这样子发现会wa（如下面的代码）
我用这个方法提交了无数次到现在还没有头绪。。。。（如果哪位大佬知道，也请一定要告诉我）
本题答案多解，我们干脆把直接把题中的N定成50再根据50来求解。然后根据逆向思维，想象再结束操作之后50个数字都是49，然后逆推，每次操作即相当于某个数加
50，其他数-1。

for(int i=1;i<=49;i++)a[i]=a[i]-m；a[50]+=m*50;                              //一直错误的代码