#include #define MAX_N 100001 #define INF_INT 2147483647 #define INF_LL 9223372036854775807 #define REP(i,n) for(int i=0;i<(int)(n);i++) using namespace std; typedef long long int ll; typedef pair P; void init(int n); int find(int n); void unite(int x,int y); bool same(int x, int y); ll bpow(ll,ll,ll); typedef vector vec; typedef vector mat; mat mul(mat &A,mat &B); mat pow(mat A,ll n); int dx[4] = {1,0,0,-1}; int dy[4] = {0,1,-1,0}; vector primes(0); vector prime(400000,true); void erast(){ prime[0] = false; prime[1] = false; for(int i=2;i<400000;i++){ if(prime[i]){ primes.push_back(i); for(int j=i+i;j<400000;j+=i){ prime[j] = false; } } } } bool cmp_P(const P &a,const P &b){ return a.second < b.second; } int main() { erast(); ll K,N,ocnt=0,d=0; int tri[1001]; tri[0] = 0; cin >> K; REP(i,1001)tri[i+1] = tri[i]+i+1; REP(i,1001){ if(K == tri[i]){ ocnt = i+1; cout << i+1 << endl; REP(j,i+1){ printf("%d ",1); } cout << endl; return 0; } if(K < tri[i]){ ocnt = i; K -= tri[i-1]; break; } } vector ss(0); ss.push_back(8); if((K) % 2 == 0){ ss.push_back(8); d++; REP(i,K/2){ ss.push_back(9); d++; } }else{ ss.push_back(8); d++; REP(i,K/2){ ss.push_back(9); d++; } ss.push_back(17); d++; ss.push_back(24); d++; } cout << ocnt+d+1 << endl; REP(i,ocnt)cout << 1 << " "; REP(i,ss.size())cout << ss[i] << " "; cout << endl; return 0; } int par[MAX_N]; int ranks[MAX_N]; //n要素で初期化 void init(int n){ REP(i,n){ par[i] = i; ranks[i] = 0; } } //木の根を求める int find(int x){ if(par[x] == x){ return x; }else{ return par[x] = find(par[x]); } } void unite(int x,int y){ x = find(x); y = find(y); if(x == y) return ; if(ranks[x] < ranks[y]){ par[x] = y; }else{ par[y] = x; if(ranks[x] == ranks[y]) ranks[x]++; } } bool same(int x, int y){ return find(x) == find(y); } ll bpow(ll a, ll n,ll mod){ int i = 0; ll res=1; while(n){ if(n & 1) res = (res*a) % mod; a = (a*a) % mod; n >>= 1; } return res; } const int MOD = 1000000007; mat mul(mat &A, mat &B){ mat C(A.size(),vec(B[0].size())); REP(i,A.size())REP(k,B.size())REP(j,B[0].size()){ C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD; } return C; } mat pow(mat A,ll n) { mat B(A.size(),vec(A.size())); REP(i,A.size()){ B[i][i] = 1; } while(n > 0){ if ( n & 1) B = mul(B,A); A = mul(A,A); n >>= 1; } return B; }