题目描述

Farmer John's N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height H_i (1 <= H_i <= 1,000,000).

Each cow is looking to her left toward those with higher index numbers. We say that cow i 'looks up' to cow j if i < j and H_i < H_j. For each cow i, FJ would like to know the index of the first cow in line looked up to by cow i.

Note: about 50% of the test data will have N <= 1,000.

约翰的N(1≤N≤10^5)头奶牛站成一排，奶牛i的身高是Hi(l≤Hi≤1,000,000)．现在，每只奶牛都在向左看齐．对于奶牛i，如果奶牛j满足i<j且Hi<Hj，我们可以说奶牛i可以仰望奶牛j． 求出每只奶牛离她最近的仰望对象．

Input

输入输出格式

输入格式：

• Line 1: A single integer: N

• Lines 2..N+1: Line i+1 contains the single integer: H_i

输出格式：

• Lines 1..N: Line i contains a single integer representing the smallest index of a cow up to which cow i looks. If no such cow exists, print 0.

输入输出样例

6

3

2

6

1

1

2

3

3

0

6

6

0

说明

FJ has six cows of heights 3, 2, 6, 1, 1, and 2.

Cows 1 and 2 both look up to cow 3; cows 4 and 5 both look up to cow 6; and cows 3 and 6 do not look up to any cow.

分析：和前几题差不多，不过就是要记录一下每次栈顶的位置.

#include <cstdio>

#include <string>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

int n,h[],ans[],stk[],top,num[];

int main()

{

    scanf("%d", &n);

    for (int i = ; i <= n; i++)

        scanf("%d", &h[i]);

    for (int i = n; i >= ; i--)

    {

        while (top !=  && stk[top] <= h[i])

            top--;

        ans[i] = num[top];

        stk[++top] = h[i];

        num[top] = i;

    }

    for (int i = ; i <= n; i++)

        printf("%d\n", ans[i]);

    return ;

}