Text Reverse
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13021 Accepted Submission(s): 4954
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.
以前不知道頭檔案<cstring>有strrev這個函數（使字元串逆向），看别人代碼我頓時無語了，尼瑪
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
const int MAX=101;
char s[MAX][MAX];
char k[MAX][MAX];
int p[MAX];
using namespace std;
int main()
{
int t,n,i,j,a,b,q;
while(cin>>t)
{
getchar();
for(i=0;i<t;i++)
{
gets(s[i]);
a=0;
b=0;
n=strlen(s[i]);
for(j=0;j<n;j++)
{
if(s[i][j]==' ')
{
p[a]=b;
a+=1;
b=0;
}
else
{
k[a][b]=s[i][j];
b+=1;
}
}
p[a]=b;
for(q=0;q<=a;q++)
{
for(j=p[q]-1;j>=0;j--)
{
cout<<k[q][j];
}
if(q!=a)
cout<<' ';
}
cout<<endl;
}
}
return 0;
}
/*-------------别人代碼--------------------
#include<stdio.h>
#include<string.h>
int main()
{
int t;
scanf("%d\n",&t);
while(t--)
{
char s[1005]={};
while(scanf("%[^ \n]",s)!=EOF)
{
printf("%s",strrev(s));
s[0]=0;
putchar(getchar());
}
}
return 0;
}
-------------------------------------------*/
坑啊！！！！