HDOJ 1062 Text Reverse

Text Reverse

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20644    Accepted Submission(s): 7887

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

 

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#include<stdio.h>
#include<string.h>
void tran(char a[1000]){
	int len,i;
	char temp;
	len=strlen(a);
	for(i=0;i<len/2;i++){ //一定是i<len/2，切记 
		temp=a[i]; a[i]=a[len-i-1]; a[len-i-1]=temp;
	}
}      //翻转 
int main(){
	int n,i,len,t;
	char a[1000],b[1000];
	scanf("%d",&n);
	getchar();  //接收 \n 
	while(n--){
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		gets(a);
		len=strlen(a);
		t=0;
		for(i=0;;i++){   //终止条件是 i==len-1
			if(a[i]!=' '){
				b[i-t]=a[i];
			}  //在不等于字符串长度的情况下，遇到空格就是一个子串的结束 
			else {
				tran(b);
				printf("%s ",b);
				t=i+1;
				memset(b,0,sizeof(b));
			}
			if(i==len-1){
				tran(b);
				printf("%s\n",b);
				memset(b,0,sizeof(b));
				break;
			}	// 等于字符串长度就是一个子串的结束 
		}
	}
	return 0;
}