AtCoder Regular Contest 079-D

D - Decrease (Contestant ver.)

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:

N
a1 a2 … aN
Here, 2≤N≤50 and 0≤ai≤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

题目大意：有一个数列{an}，给出操作个数k，每次操作选取最大的数减N，其余数+1，操作结束后，该数列中最大的数<=n-1，问n和该数列为多少。
解题思路：逆向思维，设原数列为{0,1,2,…,49}，从左到右的数即为选择的数。注意边界大小。

#include<iostream>#include<cstdio>#include<vector>#include<map>#include<set>#include<queue>#include<cmath>#include<string>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const LL INF=1e18;const int MAXN=2e5+100;const double eps=1e-10;int main(){    LL k,n=50;    while(scanf("%lld",&k)!=EOF)    {        printf("50\n");        LL cnt=k/50,res=k%50;        for(int i=0;i<50;i++)        {            if(i<res) printf("%lld",i+cnt+n-res+1);            else printf("%lld",i+cnt-res);            if(i!=49) printf(" ");        }        printf("\n");    }    return 0;}