hdu——1062Text Reverse（反转函数reverse）

最新推荐文章于 2024-07-28 10:45:00 发布

原创最新推荐文章于 2024-07-28 10:45:00 发布 · 548 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #hdu

编程算法专栏收录该内容

75 篇文章

订阅专栏

本文介绍了一个字符串反转的问题，挑战在于反转由单行文本组成的多个单词。文章提供了完整的代码示例，展示了如何通过查找和反转每个单词来解决问题，并给出了样例输入和输出。

Text Reverse
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25706 Accepted Submission(s): 9959

Problem Description

Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single line with several words. There will be at most 1000 characters in a line.

Output

For each test case, you should output the text which is processed.

Sample Input

3
olleh !dlrow
m'I morf .udh
I ekil .mca

Sample Output

hello world!
I'm from hdu.
I like acm.

Hint

Remember to use getchar() to read '\n' after the interger T, then you may use gets() to read a line and process it.

Author

Ignatius.L

Recommend

注意最后一部份也要反转

#include<string>
#include<algorithm>
#include<iostream>
#include<sstream>
using namespace std;
int main()
{
	long long m,n;
	string a;
	cin>>n;
	getchar();
	while(n--)
	{
		m=0;
		getline(cin,a);
		while(a.find(" ",m)!=string::npos)
		{
			reverse(a.begin()+m,a.begin()+a.find(" ",m));
			m=a.find(" ",m)+1;
		}
		reverse(a.begin()+a.find_last_of(" ")+1,a.end());
		cout<<a<<endl;
	}
	return 0;
}