CF1156E Special Segments of Permutation（单调栈）

题目

给定一个长度为$n$的排列$p$，求有多少区间$[l,r]$满足，$p[l]+p[r]=maxp[i]$，其中$l<=i<=r$

题解

• 预处理出左边第一个大于$i$的数和右边第一个大于$i$的数(单调栈维护一下即可)
• 左端点显然只能在$(L,i)$，右端点只能在$(i,R)$，枚举长度较小的区间判断是否可行即可

code

#include 

using namespace std; 

const int N = 2e5 + 100; 

int n, top; 
int a[N], s[N], L[N], R[N], pos[N]; 

int main() {
	scanf("%d", &n); 
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]); 
		pos[a[i]] = i; 
	}
	s[top = 0] = 0; 
	for (int i = 1; i <= n; ++i) {
		while (top && a[i] > a[s[top]]) top--; 
		L[i] = s[top], s[++top] = i; 
	}
	s[top = 0] = n + 1; 
	for (int i = n; i >= 1; --i) {
		while (top && a[i] > a[s[top]]) top--; 
		R[i] = s[top], s[++top] = i; 
	}
	int ans = 0; 
	for (int i = 1; i <= n; ++i) {
		if (i - L[i] < R[i] - i) {
			int Max = a[i]; 
			for (int j = L[i] + 1; j < i; ++j) {
				int tmp = a[j]; 
				if (pos[Max - tmp] > i && pos[Max - tmp] < R[i]) ++ans; 
			}
		}
		else {
			int Max = a[i]; 
			for (int j = i + 1; j < R[i]; ++j) {
				int tmp = a[j]; 
				if (pos[Max - tmp] > L[i] && pos[Max - tmp] < i) ++ ans; 
			}
		}
	}
	printf("%d\n", ans);
	return 0; 
}