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Text Reverse

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26663    Accepted Submission(s): 10393 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.

虽然是道水题，但是有好多需要注意的地方，就贴一下

#include<cstdio> #include<cstring> #include<iostream> using namespace std; char str[1010]; int main() { int t; scanf("%d",&t); getchar(); while(t--) { gets(str); int len=strlen(str); int start=0; bool flag=0; for(int i=0;i<len;i++) { if(str[i]==' ') { if(flag) printf(" "); for(int k=i-1;k>=start;k--) printf("%c",str[k]); start=i+1; flag=1; } if(i==len-1) { printf(" "); for(int k=i;k>=start;k--) printf("%c",str[k]); } } puts(""); /* for(int i=0;i<len;i++) { if(str[i]==' '||i==len-1) 这样写是 PE的，举个特例：a_b_ 这个程序的答案是：a__b,正确答案应该是：a_b { if(flag) printf(" "); for(int k=(i==len-1?i:i-1);k>=start;k--) printf("%c",str[k]); start=i+1; flag=1; } } puts("");*/ } return 0; }