题面说的有点问题，应该是向后看齐。
于是我们维护一个单调递减栈，如果当前a[i]比栈顶元素大，就执行pop操作，然后把pop出来的元素的答案都用 i 更新即可。
1 #include<cstdio>
2 #include<iostream>
3 #include<cmath>
4 #include<algorithm>
5 #include<cstring>
6 #include<cstdlib>
7 #include<cctype>
8 #include<vector>
9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(' ')
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const db eps = 1e-8;
19 const int maxn = 1e5 + 5;
20 inline ll read()
21 {
22 ll ans = 0;
23 char ch = getchar(), last = ' ';
24 while(!isdigit(ch)) {last = ch; ch = getchar();}
25 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
26 if(last == '-') ans = -ans;
27 return ans;
28 }
29 inline void write(ll x)
30 {
31 if(x < 0) x = -x, putchar('-');
32 if(x >= 10) write(x / 10);
33 putchar(x % 10 + '0');
34 }
35
36 int n, a[maxn];
37 #define pr pair<int, int>
38 #define mp make_pair
39 stack<pr> st;
40 int ans[maxn];
41
42 int main()
43 {
44 n = read();
45 for(int i = 1; i <= n; ++i) a[i] = read();
46 for(int i = 1; i <= n; ++i)
47 {
48 while(!st.empty() && st.top().first < a[i])
49 {
50 pr node = st.top(); st.pop();
51 ans[node.second] = i;
52 }
53 st.push(mp(a[i], i));
54 }
55 for(int i = 1; i <= n; ++i) write(ans[i]), enter;
56 return 0;
57 }
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