Sum of Three Terms

题意

ARC135B
给出 \(n\) 个数的数列 \(S\)，求是否存在长度是 \(n + 2\) 的数列 \(A\), 满足 \(\forall i \in [1, n], S_i = A_i + A_{i + 1} + A_{i + 2}\), \(\forall i \in [1, n + 2], A_i \geq 0\)。

analysis

\[A_{i + 2} = S_i - A_{i + 1} - A_{i} \]

当给定 \(A_{1}, A{2}\) 时，就能计算出整个数列。

考虑待定系数 \(a = A_{1}, b = A_{2}\)。

带入上面的运算，则有以下性质:

\[\left\{\begin{aligned} A_i &= x_i + a & (i \equiv 1 \pmod 3) \\ A_i &= x_i + b & (i \equiv 2 \pmod 3) \\ A_i &= x_i - a - b & (i \equiv 0 \pmod 3) \end{aligned}\right. \]

其中 \(x\) 是原来的数列 \(A\)。

它们可以转化为不等式组：

\[\left\{\begin{aligned} a \geq - x_i & (i \equiv 1 \pmod 3) \\ b \geq - x_i & (i \equiv 2 \pmod 3) \\ x_i \leq a + b & (i \equiv 0 \pmod 3) \end{aligned}\right. \]

因此解出这个不等式组即可。

令 \(S_1 = \max_{i \equiv 1 \pmod 3} - x_i, S_2 = \max_{i \equiv 2 \pmod 3} - x_i, S_3 = \min_{i \equiv 0 \pmod 3} x_i\)

如果 \(S_1 + S_2 > S_3\) 即无解。
反之，\((A_1, A_2) = (S_1, S_2)\)。

代码

#include<bits/stdc++.h>
using namespace std;

using ll = long long;
const int MAXN = 300010;
const int INF = 0x7fffffff;
const int mod = 1000000007;

template <typename T>
void Read(T &x) {
	x = 0; T f = 1; char a = getchar();
	for(; a < '0' || '9' < a; a = getchar()) if (a == '-') f = -f;
	for(; '0' <= a && a <= '9'; a = getchar()) x = (x * 10) + (a ^ 48);
	x *= f;
}

int n, len;
ll S[MAXN], x[MAXN];
int main() {
	cin >> n;
	for (int i = 1; i <= n; i ++) {
		cin >> S[i];
		x[i + 2] = S[i] - x[i + 1] - x[i];  
	}
	ll S1 = -INF, S2 = -INF, S3 = INF;
	for (int i = 1; i <= n + 2; i ++) {
		if (i % 3 == 1) S1 = max(S1, - x[i]);
		if (i % 3 == 2) S2 = max(S2, - x[i]);
		if (i % 3 == 0) S3 = min(S3, x[i]);
	}
	if (S1 + S2 > S3) 
		return cout << "No", 0;
	cout << "Yes" << endl;
	ll a = S1, b = S2;
	for (int i = 1; i <= n + 2; i ++) {
		if (i % 3 == 1) cout << x[i] + a << ' ';
		if (i % 3 == 2) cout << x[i] + b << ' ';
		if (i % 3 == 0) cout << x[i] - a - b << ' '; 
	}
	return 0;
}