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  • PAT 甲级 1051 Pop Sequence

    https://pintia.cn/problem-sets/994805342720868352/problems/994805427332562944

    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

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int N, M, K;
    vector<int> v;
    
    int main() {
        scanf("%d%d%d", &M, &N, &K);
        while(K --) {
            bool flag = false;
            v.resize(N + 1);
            int cnt = 1;
            for(int i = 1; i <= N; i ++)
                scanf("%d", &v[i]);
    
            stack<int> s;
            for(int i = 1; i <= N; i ++) {
                s.push(i);
                if(s.size() > M) break;
                while(!s.empty() && s.top() == v[cnt]) {
                    s.pop();
                    cnt ++;
                }
            }
    
            if(cnt == N + 1) flag = true;
            if(flag) printf("YES
    ");
            else printf("NO
    ");
        }
        return 0;
    }
    

      感觉年就这么结束了诶

     

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10354106.html
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