[USACO09MAR]Look Up

嘟嘟嘟

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。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a) memset(a, 0, sizeof(a))
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e5 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n, a[maxn];
 #define pr pair<int, int>
 #define mp make_pair
 stack<pr> st;
 int ans[maxn];

 int main()
 {
     n = read();
     for(int i = ; i <= n; ++i) a[i] = read();
     for(int i = ; i <= n; ++i)
     {
         while(!st.empty() && st.top().first < a[i])
         {
             pr node = st.top(); st.pop();
             ans[node.second] = i;
         }
         st.push(mp(a[i], i));
     }
     for(int i = ; i <= n; ++i) write(ans[i]), enter;
     return ;
 }