zoukankan      html  css  js  c++  java
  • PAT 甲级 1057 Stack

    https://pintia.cn/problem-sets/994805342720868352/problems/994805417945710592

    Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian -- return the median value of all the elements in the stack. With N elements, the median value is defined to be the (-th smallest element if N is even, or (-th if N is odd.

    Input Specification:

    Each input file contains one test case. For each case, the first line contains a positive integer N (≤). Then N lines follow, each contains a command in one of the following 3 formats:

    Push key
    Pop
    PeekMedian
    

    where key is a positive integer no more than 1.

    Output Specification:

    For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print Invalid instead.

    Sample Input:

    17
    Pop
    PeekMedian
    Push 3
    PeekMedian
    Push 2
    PeekMedian
    Push 1
    PeekMedian
    Pop
    Pop
    Push 5
    Push 4
    PeekMedian
    Pop
    Pop
    Pop
    Pop
    

    Sample Output:

    Invalid
    Invalid
    3
    2
    2
    1
    2
    4
    4
    5
    3
    Invalid

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1e5 + 10;
    int N;
    int c[maxn];
    stack<int> s;
    
    int lowerbit(int x) {
        return x & -x;
    }
    
    void update(int x, int v) {
        for(int i = x; i < maxn; i += lowerbit(i))
            c[i] += v;
    }
    
    int getsum(int x) {
        int sum = 0;
        for(int i = x; i >= 1; i -= lowerbit(i))
            sum += c[i];
        return sum;
    }
    
    void PeekMedian() {
        int left = 1, right = maxn, mid, k = (s.size() + 1) / 2;
        while(left < right) {
            mid = (left + right) / 2;
            if(getsum(mid) >= k)
                right = mid;
            else left = mid + 1;
        }
        printf("%d
    ", left);
    }
    
    int main() {
        scanf("%d", &N);
        while(N --) {
            string op;
            cin >> op;
            if(op == "Pop") {
                if(s.empty()) printf("Invalid
    ");
                else {
                    printf("%d
    ", s.top());
                    update(s.top(), -1);
                    s.pop();
                }
            } else if(op == "Push") {
                int num;
                scanf("%d", &num);
                s.push(num);
                update(num, 1);
            } else {
                if(s.empty()) printf("Invalid
    ");
                else PeekMedian();
            }
        }
        return 0;
    }
    

      树状数组 第一次写到树状数组 看博客还没看的很懂但是隐约感觉到这是一个模板题 emmmm 以后多做一点就会明白了叭 唉 哭唧唧

  • 相关阅读:
    python基础之流程控制
    多线程---阻塞队列
    多线程---线程同步
    多线程---线程实现
    多线程start();之后会开辟新的栈空间
    java中使用String的split分隔字符串注意事项
    IO流
    java中的多态
    关于java中的接口
    关于final关键字
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10357628.html
Copyright © 2011-2022 走看看