HDU 1062 Text Reverse 水题 字符串处理

题意：把一行文本中以空格间隔的单词反向。

思路：遇到空格就处理即可。

http://acm.hdu.edu.cn/showproblem.php?pid=1062

/*********************************************    Problem : HDU 1062    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 1e5+5;const int MAXE = 2e5+5;typedef long long LL;typedef unsigned long long ULL;int T,n,m,k;int fx[] = {0,1,-1,0,0};int fy[] = {0,0,0,-1,1};char s[1005];void input() {    gets(s);}void reverse(int s1,int e1) {    char tmp;    for(int i = s1 , j = e1 ; i < j ; i ++ , j --) {        tmp = s[i];        s[i] = s[j];        s[j] = tmp;    }}void solve() {    int lens = strlen(s);    int s1 = 0; int e1 ;    rep(i,0,lens-1) {        if(s[i] == ' ') {            e1 = i - 1;            reverse(s1,e1);            //printf("%d %d",s1,e1);            s1 = i + 1;        }        else if(i == lens-1) {            e1 = i ;            reverse(s1,e1);            //printf("%d %d",s1,e1);        }    }    puts(s);}int main(void) {    //freopen("a.in","r",stdin);    scanf("%d",&T); getchar();    while(T--) {    //while(~scanf("%d %d",&n,&m)) {    //while(scanf("%d",&n),n) {        input();        solve();    }    return 0;}