题意
Sol
对构造一无所知。。。
题解的方法比较神仙，，设第一个位置为\(-1\)，\(S = \sum_{i=1}^n a_i\)
那么我们要让\(N * S - (N - 1) * (S + 1) = K\)
固定\(N\)之后可以直接解出\(S\)。。。
#include<bits/stdc++.h>
#define Pair pair<LL, LL>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 2001, mod =666623333, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N = 2000, a[MAXN];
signed main() {
a[1] = -1;
int Lim = 1e6, K = read(), S = K + N;
for(int i = 2; i <= N; i++)
if(S > Lim) a[i] = Lim, S -= Lim;
else {a[i] = S; break;}
cout << N << '\n';
for(int i = 1; i <= N; i++) cout << a[i] << " ";
return 0;
}