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 yppa yadot
OT eb ro ton ot eb
I tnaw ot niw eht ecitcarp tsetnoc
Analysis:
I personally feel that the knowledge of using Stacks is better...
Today, I have reviewed my previous knowledge and I feel that I have forgotten a lot of my previous knowledge. But now I will review it, remember it quickly, and use it.
Source code:
#include<stdio.h> #include<algorithm> #include<iostream> #include<stack> #include<string.h> using namespace std; stack<char>z; int main() { int k; int n; char a[999]; scanf("%d",&k); while(k--) { scanf("%d",&n); getchar(); while(n--) { gets(a); int l=strlen(a); int i=0; while(1) { while(a[i]!=' '&&i<l) { z.push(a[i]); i++; } while(!z.empty()) { printf("%c",z.top()); z.pop(); } if(i!=l) { printf(" "); i++; } else break; } printf("\n"); } if(k) printf("\n"); } return 0; } #include<stdio.h>#include<algorithm>#include<iostream>#include<stack>#include<string.h>using namespace std;stack<char>z;int main(){ int k; int n; char a[999]; scanf("%d",&k); while(k--) { scanf("%d",&n); getchar(); while(n--) { gets(a); int l=strlen(a); int i=0; while(1) { while(a[i]!=' '&&i<l) { z.push(a[i]); i++; } while(!z.empty()) { printf("%c",z.top()); z.pop(); } if(i!=l) { printf(" "); i++; } else break; } printf("\n"); } if(k) printf("\n"); } return 0;}