大意: 求將[1,n]劃分成兩個集合, 且兩集合的和的差盡量小.

和/2為偶數最小差一定為0, 和/2為奇數一定為1.

顯然可以通過某個前綴和刪去一個數得到.

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, a[N];
char s[N];



int main() {
	scanf("%d", &n);
	int s = (1+n)*n/2, t = 0;
	REP(i,1,n) {
		t += i;
		if (t>=s/2&&t-s/2<=i) {
			vector<int> g;
			int r = 0;
			REP(j,1,i) if (j!=t-s/2) g.pb(j),r+=j;	
			int rr = s-r;
			printf("%d\n%d ", abs(r-rr),int(g.size()));
			for (auto &&t:g) printf("%d ",t);
			return hr,0;
		}
	}
}

Dividing the numbers CodeForces - 899C (構造)