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
#include <bits/stdc++.h> using namespace std; int main(){ int m,n,k; cin>>m>>n>>k; while(k--){ stack<int> s; int i=1; int a[1005],f=0; for(int j=0;j<n;j++){ cin>>a[j]; if(f)continue; while(i<=n+1){ if(s.size()==m){ if(s.top()!=a[j])f=1; else s.pop(); break; } else if(s.size()<m){ if(s.empty())s.push(i++); else{ if(s.top()==a[j]){ s.pop();break; } else if(s.top()<a[j]) s.push(i++); else if(s.top()>a[j]){ f=1;break; } } } } } if(f)cout<<"NO"<<endl; else{ if(s.empty())cout<<"YES"<<endl; else cout<<"NO"<<endl; } } }
卡了一会儿,出现错误现象:数据没输完却输出了完整数量的答案,找了一会错,发现自己f=1后立刻将整个程序break掉了,没让数据输完。