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.
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.HintRemember 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
法一
#include<stdio.h> #include<string.h> int main() { int i,n,len,j,k,t; char s1[1005],s2[100]; scanf("%d",&n); getchar(); while(n--) { gets(s1); len=strlen(s1); for(i=0,j=0,t=0;i<len;i++) { if(s1[i]!=' ') s2[j++]=s1[i]; /*保存单词*/ else { if(t>0) printf(" "); /*控制格式*/ for(k=j-1;k>=0;k--) printf("%c",s2[k]); /*反转输出*/ j=0; t++; } if(i==len-1) /*反转最后一个单词,这里要特别注意*/ { printf(" "); for(k=j-1;k>=0;k--) printf("%c",s2[k]); } } printf("\n"); } return 0; }思路:定义两个数组,一个输入,然后存入另一个数组反转输出即可。回过头来看当时写的代码真的好麻烦,所以有写了一个只用一个代码的。
#include<stdio.h>
#include<string.h>
int main()
{
int t,f,i,j,k;
char s[1005];
scanf("%d",&t);
getchar();
while(t--)
{
gets(s);
f=strlen(s);
for(i=0; i<=f; i++)
{
if(s[i]==' '||i==f)
{
for(j=i-1; j!=k&&j>=0; j--)
putchar(s[j]);
k=i;
if(i!=f)
putchar(' ');
}
}
printf("\n");
}
}
然后我又发现了反转函数,其实还可以写一个的再写一个: ),有时间了再写吧。
原型:extern char *strrev(char *s);
用法:#include <
string.h>
功能:把字符串s的所有字符的顺序颠倒过来(不包括空字符NULL)。
说明:返回指向颠倒顺序后的字符串
指针。