uva 11988 Broken Keyboard (a.k.a. Beiju Text)
2013-07-26 12:41
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点击打开链接uva 11988
思路: deque模拟
分析:
1 题目给定一个字符串要求通过一序列的模拟输出最后的字符串
2 根据题目的意思[],分别表示的是键盘上的home和end键,home键的作用是跳到起始位置,end的作用是到最后一个位置。
3 根据2我们可以利用双端队列来模拟,如果是[我们插入front,如果是]插入back,最后在输出即可
代码:
list 来做
#include<list>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 100010;
list<char>ls;
int main(){
char str[MAXN];
while(gets(str)){
ls.clear();
int len = strlen(str);
list<char>::iterator it = ls.begin();
for(int i = 0 ; i < len ; i++){
if(str[i] == '[')
it = ls.begin();
else if(str[i] == ']')
it = ls.end();
else{
ls.insert(it,str[i]);
}
}
for(it = ls.begin(); it != ls.end() ; it++)
printf("%c" , *it);
puts("");
}
return 0;
}
思路: deque模拟
分析:
1 题目给定一个字符串要求通过一序列的模拟输出最后的字符串
2 根据题目的意思[],分别表示的是键盘上的home和end键,home键的作用是跳到起始位置,end的作用是到最后一个位置。
3 根据2我们可以利用双端队列来模拟,如果是[我们插入front,如果是]插入back,最后在输出即可
代码:
#include<deque> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int MAXN = 100010; char str[MAXN]; deque<string>dqe; void insert(bool front , bool rear , string s){ if(front) dqe.push_front(s); if(rear) dqe.push_back(s); } void output(){ while(!dqe.empty()){ cout<<dqe.front(); dqe.pop_front(); } puts(""); } void solve(){ int len = strlen(str); bool front , rear; string s = ""; front = true; rear = false; for(int i = 0 ; i < len ; i++){ if(str[i] == '['){ insert(front , rear , s); s = ""; front = true; rear = false; } else if(str[i] == ']'){ insert(front , rear , s); s = ""; front = false; rear = true; } else{ s += str[i]; } } insert(front , rear , s); output(); } int main(){ while(scanf("%s" , str) != EOF) solve(); return 0; }
list 来做
#include<list>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 100010;
list<char>ls;
int main(){
char str[MAXN];
while(gets(str)){
ls.clear();
int len = strlen(str);
list<char>::iterator it = ls.begin();
for(int i = 0 ; i < len ; i++){
if(str[i] == '[')
it = ls.begin();
else if(str[i] == ']')
it = ls.end();
else{
ls.insert(it,str[i]);
}
}
for(it = ls.begin(); it != ls.end() ; it++)
printf("%c" , *it);
puts("");
}
return 0;
}
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