Sum of three terms (construction)

https://atcoder.jp/contests/arc135/tasks
#include<iostream>
#include<map>
#include<cstring>
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 pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
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');
}
const int mod = 998244353;
const int N = 3e5 + 10;

int a[N];
int s[N];
int n;
void solve() {
	cin >> n;
	for (int i = 0; i< n; i ++) cin >> a[i];
	
	int minvy = 0, minvx = 0;
	// Negative numbers need not be considered , Because he was a positive number , It must be greater than or equal to 0 Conditions , Then we just need to find the smallest negative number 
	int maxx = 0x3f3f3f3f;
	
	for (int i = 2; i< n + 2; i ++) {
		s[i] = a[i - 2] - s[i - 1] - s[i - 2];
		if (i % 3 == 0) {
			minvy = max (minvy, -s[i]);
		} 
		else if (i % 3 == 1) {
			minvx = max(minvx, -s[i]);
		}
		else {
			maxx  = min(maxx, s[i]);
		}
	}
	if (minvy + minvx > maxx) {
		puts("No");
		return ;
	}
	puts("Yes");
	s[0] = minvy, s[1] = minvx;
	for (int i = 2; i < n + 2; i ++) 
		s[i] = a[i - 2] - s[i - 1] - s[i - 2];
	for (int i = 0; i < n + 2; i ++)
		cout << s[i] << " \n"[i == n + 1];
}
int main () {
    int t;
    t = 1;
    while (t --) solve();
    return 0;
}

This problem is solved by finding the boundary range and then constructing a feasible solution . By simplifying the formula You can find i % 3 == 0 s[i] = x +a i % 3 == 1 s[i] = x + b i % 3== 2 s[i] = x - a - b We need to find out whether the limit conditions are feasible To this end, we construct c1, c2, c3 c1 <= a c2 <= b a + b <=c3 c1 + c2<= a + b <= c3 Maximum left , Take the minimum on the right and judge whether it meets the size relationship to judge whether it is feasible . If it is satisfied, just substitute the value directly