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    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 N1 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 N1 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
    • 0K50×1016
    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, 2N50 and 0ai1016+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
    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 */

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  • 原文地址:https://www.cnblogs.com/lu1nacy/p/10013822.html
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