joj 1146 标准输入+字符串反转

1146: Word Reversal

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Result TIME Limit MEMORY Limit Run Times AC Times JUDGE

	3s	8192K	2445	640	Standard

For each list of words, output a line with each word reversed without changing the order of the words.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

You will be given a number of test cases. The first line contains a positive integer indicating the number of cases to follow. Each case is given on a line containing a list of words separated by one space, and each word contains only uppercase and lowercase letters.

Output

For each test case, print the output on one line.

Sample Input

1

3
I am happy today
To be or not to be
I want to win the practice contest

Sample Output

I ma yppah yadot
oT eb ro ton ot eb
I tnaw ot niw eht ecitcarp tsetnoc

 

/*
关键的不是算法
而是怎么读入一行中间有空格的字符串

  算法：
		有一个空格，就把tmp中的内容反转
		
*/
#include <iostream>
#include <cstring>

using namespace std;
char str[500];
char tmp[50];
int n,m;
int flag = 0;
int main()
{
	freopen("in.txt","r",stdin);
	cin >> m;
	while(m-- >0)
	{
		if(flag = 0)
			cout<<endl;
		else
			flag = 1;
		cin >> n;
		cin.get () ;//这个函数是必须的，否则换行之类的没有处理
		for(int i=0;i<n;i++)
		{
			memset(str,'\0',sizeof(str));
			memset(tmp,'\0',sizeof(tmp));
			//gets(str); //破函数，啥都读不进去
			cin.getline (str, 500, '\n') ; //函数的使用
			int j =0;
			for(int i=0;i<strlen(str);i++)
			{
				
				if(str[i] != ' ')
				{
					tmp[j] = str[i];
					j ++;
				}
				else
				{
					j -= 1;
					while(j > -1) //反向赋值
					{
						str[i - j - 1] = tmp[j];
						j -- ;
					}
					j = 0;
				}
				if(strlen(str) -1 == i)
				{
					j -= 1;
					while(j > -1) //反向赋值
					{
						str[i - j] = tmp[j];
						j -- ;
					}
					j = 0;
				}
			}
			cout << str<<endl;
		}
	}
	fclose(stdin);
	return 0;
}

 

就是一道水题，然后标准输入函数的使用

JOJ上PE了。算法没有问题，果断放弃