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 }