[Swust OJ 1026]--Egg pain's hzf

最新推荐文章于 2022-06-15 12:41:24 发布

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最新推荐文章于 2022-06-15 12:41:24 发布

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原文链接：http://www.cnblogs.com/zyxStar/p/4542800.html

本文详细介绍了如何使用C++容器解决在一组整数中找到每个数左侧比其小的最大数的问题。通过引入set容器，并利用其lower_bound函数，实现了高效的查找逻辑。同时，文章提供了一个实例输入输出示例，帮助读者理解实现过程。

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题目链接： http://acm.swust.edu.cn/problem/1026/

Time limit(ms): 3000　　　　Memory limit(kb): 65535

hzf is crazy about reading math recently,and he is thinking about a boring problem.

Now there are n integers Arranged in a line.For each integer,he wants to know the maximum integer in its left which is less than it.

Description

The input consists of multiple test cases.

For each case,the first line contains one integer n（0<n<=100000）,indicating the number of integers.

The second line contains n integers,indicating the integers from left to right,marked a1…an.(0<ai<=10^9).

Input

For each case,there are n integers,indicating the maximum integer in ai's left which is less than it.If it doesn's exist,output -1.

Output

Sample Input

1 2	-1 1 2 3 4 -1 -1 -1 -1 -1

Sample Output

由于OJ上传数据的BUG,换行请使用"\r\n"，非常抱歉

题目大意：给你一排数字，在这个状态下，在当前数的左边找比它小的最大数字，没有就是-1

思路：这个题说白了就是一个容器set的用法（当然其他方法也可以），这个头文件类有点多
给个知识链接：http://www.cnblogs.com/zyxStar/p/4542835.html代码如下

 1 #include <iostream>
 2 #include <set>
 3 using namespace std;
 4 int n, x[100001];
 5 int main(){
 6     while (cin >> n){
 7         set<int>mpt;
 8         set<int>::iterator it;
 9         int i, k;
10         for (i = 0; i < n; i++){
11             cin >> x[i];
12             k = x[i];
13             it = mpt.lower_bound(x[i]);
14             if (it == mpt.begin()) x[i] = -1;
15             else{
16                 it--;
17                 x[i] = *it;
18             }
19             mpt.insert(k);
20         }
21         for (i = 0; i < n; i++){
22             if (i) cout << ' ';
23             cout << x[i];
24         }
25         cout << "\r\n";
26     }
27     return 0;
28 }