1050 String Subtraction （20 point(s)）
Given two strings S1 and S2, S=S1−S2 is defined to be the remaining string after taking all the characters in S2 from S1. Your task is simply to calculate S1−S2 for any given strings. However, it might not be that simple to do it fast.
Input Specification:
Each input file contains one test case. Each case consists of two lines which gives S1 and S2, respectively. The string lengths of both strings are no more than 104. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.
Output Specification:
For each test case, print S1−S2 in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.题目大意：给两个字符串，给出S1-S2，其中减法规则是把前者里面在后者字符串中存在的字符全部删去。
解题思路：However, it might not be that simple to do it fast.别被这句话吓到。。其实很水。
用set存一下第二个字符串，然后对第一个字符串进行遍历，直接用STL的find函数就行了，如果在集合里找得到，就不输出，找不到就输出。
代码如下：
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
set<char> m;
int main()
{
string a, b;
getline(cin, a);
getline(cin, b);
for (int i = 0; i < b.size(); i++)m.insert(b[i]);
for (int i = 0; i < a.size(); i++) {
if (find(m.begin(), m.end(), a[i])==m.end()) {
cout << a[i];
}
}
cout << endl;
return 0;
}1051 Pop Sequence （25 point(s)）
Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.
Input Specification:
Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K(the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.
Output Specification:
For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.
Sample Input:
5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2
Sample Output:
YES
NO
NO
YES
NO题目大意：就是数据结构笔试经常做到的。。告诉你一个进堆栈的序列，然后给你一些出栈的序列，判断这种是否合法，注意一下这里的堆栈有大小限制，也就是堆栈满了就必须pop
解题思路：新建一个堆栈模拟一下，每次从1开始读，进栈之后判断，是不是超出限制了，如果超出了，不用继续，直接就是不合法的。如果没超出，那继续判断是不是和当前数组指针指向的数一致，如果一致，就pop，如此一直继续到不相等或者栈空即可。
代码如下：
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
#include<stack>
using namespace std;
int main()
{
int M, N, K;
cin >> M >> N >> K;
while (K--) {
stack<int> tmp;
vector<int> a;
for (int i = 0; i < N; i++) {
int c;
cin >> c;
a.push_back(c);
}
bool flag=true;
int index = 0;
for (int i = 1; i <= N; i++) {
tmp.push(i);
if (tmp.size() > M) {
flag = false;
break;
}
while(!tmp.empty() && tmp.top() == a[index]) {
tmp.pop();
index++;
}
}
if (tmp.empty()&& flag == true) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
}
return 0;
}