CF1787C - Remove the Bracket

https://codeforces.com/problemset/problem/1787/C
This is the reason why the problem was named as Remove the Bracket.

$$\begin{array}{rlrl}\text{Product}& =a_1\cdot a_2\cdot a_3\cdot \dots \cdot a_n=& =a_1\cdot (x_2+y_2)\cdot (x_3+y_3)\cdot \dots \cdot (x_{n-1}+y_{n-1})\cdot a_n=& \stackrel{\text{?}}{=}{a}_1\cdot x_2+y_2\cdot x_3+y_3\cdot \dots \cdot x_{n-1}+y_{n-1}\cdot a_n.\end{array}$$

$(x_i-s)(y_i-s)\ge 0$ tells us either $min(x_i,y_i)\ge$ s or $max(x_i,y_i)\le$ s, so pickable $x_i$ is a consecutive range.$

Just consider $(x_i+y_i)$, remove the bracket then it turns to \ldots+ y_{i-1}\cdot x_i+y_i\cdot x_{i+1}+\ldots. When y_{i-1} = x_{i+1}, the result is constant, so we assume that $y_{i-1}<x_{i+1}$.

If x_i does not hit the maximum, increase $x_i$ by 1 and decrease $y_i$ by 1, the result will increase by $y_{i-1}$ and decrease by $x_{i+1}$, which means decrease by $x_{i+1}-y_{i-1}$, always better. So $x_i$ will either hit the maximum or the minimum. Thus, each number has only two ways of rewriting that might be optimal.

This can be done with DP.

#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long f[N][2],x[N],y[N];
void get() {
	int i,n,s,j;
	cin>>n>>s;
	for(i=1; i<=n; i++) {
		cin>>j;
		if(i==1||i==n) x[i]=y[i]=j;
		else if(j<=s) x[i]=0,y[i]=j;
		else x[i]=s,y[i]=j-s;
	}
	f[1][0]=f[1][1]=0;
	for(i=2; i<=n; i++) {
		f[i][0]=min(f[i-1][0]+y[i-1]*x[i],f[i-1][1]+x[i-1]*x[i]);
		f[i][1]=min(f[i-1][0]+y[i-1]*y[i],f[i-1][1]+x[i-1]*y[i]);
	}
	cout<<f[n][0]<<endl;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int T; cin>>T;
	while(T--) get();
	return 0;
}