A 数列下标
题目地址:
基本思路:
找离一个位置最近的比他大的数的位置,比较明显的一个单调栈,
不懂单调栈的同学可以去学习一下,就是在入栈之前保证栈的单调性就可以了。
参考代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h> using namespace std; #define IO std::ios::sync_with_stdio(false) #define int long long #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 mset(s, _) memset(s, _, sizeof(s)) #define pb push_back #define pii pair <int, int> #define mp(a, b) make_pair(a, b) #define INF (int)1e18 inline int read() { int x = 0, neg = 1; char op = getchar(); while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); } while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); } return neg * x; } inline void print(int x) { if (x < 0) { putchar('-'); x = -x; } if (x >= 10) print(x / 10); putchar(x % 10 + '0'); } const int maxn = 10100; int n,a[maxn],b[maxn]; signed main() { IO; vector<int> v; cin >> n; rep(i, 1, n) cin >> a[i]; per(i, n, 1) { while (!v.empty() && a[v.back()] <= a[i]) v.pop_back(); if (!v.empty()) b[i] = v.back(); else b[i] = 0; v.push_back(i); } rep(i, 1, n) cout << b[i] << ' '; cout << '\n'; return 0; }