Description
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
Input
The single line of the input contains a sequence of characters "+" and "-" of length n (1 ≤ n ≤ 100000). The i-th (1 ≤ i ≤ n) position of the sequence contains the character "+", if on the i-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
Output
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
Sample Input
Input
-++-
Output
Yes
Input
+-
Output
No
Input
++
Output
Yes
Input
-
Output
No
Hint
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:
题意：有两根导线相互纠缠，也就是说一根导线绕一根导线，两端固定，现在想让你把他们解开———让他们相互平行不在缠绕在一起，条件是这两根到导线可以在平面上随意移动，也可以拿起放下但是不能让两端固定的交换位置，问你是否可以达到这个目标？分析：模拟简单题。。刚开始通过提示会发现只有相邻两个结点（交叉点）相同时这两个导线可以解开，于是开启智障想法，想通过对整个字符串从中间分开，对称的检查两边是否一样来解，然后就WA6，想想发现如果字符串的长度是奇数的话，以上不成立，写了一堆，GG（辣鸡），之后去网上看了提示，说用栈依次处理，和栈顶元素比较相同的元素删除栈顶，不同时进栈，最后统计栈是否为空即可。然后就发现智障选手+1.。。。刚开始的智障代码：
1 /*************************************************************************
2 > File Name: cfd.cpp
3 > Author:
4 > Mail:
5 > Created Time: 2016年07月10日 星期日 22时18分49秒
6 ************************************************************************/
7
8 #include<iostream>
9 #include<bits/stdc++.h>
10 using namespace std;
11 const int maxn = 1e5 + 7;
12 char str[maxn];
13 int main()
14 {
15 scanf("%s",str+1);
16 int len = strlen(str+1);
17 //printf("len = %d\n",len);
18 int pos = len / 2;
19 // cout << "pos = " << pos << endl;
20 bool flag = true;
21 for(int i = pos; i >= 1; i--)
22 {
23 if(str[i] != str[len+1-i])
24 {
25 flag = false;
26 break;
27 }
28 }
29 if(!flag || len == 1) printf("No\n");
30 else
31 printf("Yes\n");
32 return 0;
33 }
正确代码：
1 /*************************************************************************
2 > File Name: cfd.cpp
3 > Author:
4 > Mail:
5 > Created Time: 2016年07月10日 星期日 22时18分49秒
6 ************************************************************************/
7
8 #include<iostream>
9 #include<bits/stdc++.h>
10 using namespace std;
11 const int maxn = 1e6 + 7;
12 stack<char> s;
13 char str[maxn];
14 int main()
15 {
16 scanf("%s",str);
17 int len = strlen(str);
18 for(int i = 0; i < len; i++)
19 {
20 if(!s.empty() && str[i] == s.top())
21 {
22 s.pop();
23 }
24 else
25 {
26 s.push(str[i]);
27 }
28 }
29 if(s.empty())
30 {
31 printf("Yes\n");
32 }
33 else
34 printf("No\n");
35 return 0;
36 }