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。

/*********************************************
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-8

using 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;
}