CF

PS：换了一种方式 希望大家喜欢 2333

/**
code by: zstu wxk
time: 2019/03/01
Problem Link: http://codeforces.com/contest/1130/problem/E
solve:
    如果构造的数列是 0 0 0 0 0 0 0 -1 + + + +(+代表大于0的数)
        那么 find_answer = sum(+) * len_+
        实际上的答案为 tot_len * [sum(+) - 1]
        即: sum(+) * len_+ + k = tot_len * [sum(+) - 1]
        接下来就是枚举tot_len就好了
**/
#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
int n, k, t1, t2;
int a[N];
bool check(int x){
    for(int i = 2; i * i <= x; ++i){
        if(x % i == 0){
            t1 = i, t2 = x/i;
            t1 = n - t1;
            if(t1 < n - 2 && t2 < 1e6 * t1)
                return true;
        }
    }
    return false;
}
void Ac(){
    n = 2000;
    memset(a, 0, sizeof a);
    for(n = 2000; n >= 1; --n){
        if(check(n+k)){
            for(int i = 1; i <= t1; ++i){
                if(t2 >= 1e6){
                    a[i] = 1e6;
                    t2 -= 1e6;
                }
                else{
                    a[i] = t2;
                    t2 = 0;
                }
            }
            a[t1+1] = -1;
            printf("%d
", n);
            reverse(a+1, a+1+n);
            for(int i = 1; i <= n; ++i){
                printf("%d ", a[i]);
            }
            break;
        }
    }
}
int main(){
    while(~scanf("%d", &k)){
        Ac();
    }
    return 0;
}