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,每只奶牛都在向右看齐.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
输入格式
-
Line 1: A single integer: N
-
Lines 2..N+1: Line i+1 contains the single integer: H_i
第 1 行输入 N,之后每行输入一个身高 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.
共 N 行,按顺序每行输出一只奶牛的最近仰望对象,如果没有仰望对象,输出 0。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e5+10;
int a[N],q[N],ans[N];
int main(){
int n;
cin>>n;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
q[1]=1;
int l=1,r=1;
for(int i=1;i<=n;i++){
while(r>=l&&a[q[r]]<a[i]){
ans[q[r]]=i;
r--;
}
q[++r]=i;
}
for(int i=1;i<=n;i++)printf("%d
",ans[i]);
}