// Problem: C. Dividing the numbers
// Contest: Codeforces - Codeforces Round #452 (Div. 2)
// URL: https://codeforces.com/problemset/problem/899/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-11 19:03:19
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> Vll;
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi = acos(-1.0);
inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}
void solve() {
int n;
cin >> n;
vector<int> a;
int ans =0;
for (int i = n; i >= 1; i --) {
if (ans <= 0) {
ans += i;
a.pb(i);
}
else ans -= i;
}
cout << abs(ans) << endl;
cout << a.size() << ' ';
for (auto t : a) cout << t << ' ';
}
int main () {
// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
int t;
t =1;
//cin >> t;
while (t --) solve();
return 0;
}
维护差值。我们想要尽量靠近0,所以一旦小于0就放。从大到小贪心。类似硬币
需要很多情况
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