zoukankan      html  css  js  c++  java
  • 1057 Stack (30)

    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 (N/2)-th smallest element if N is even, or ((N+1)/2)-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 (<= 10^5^). 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 10^5^.

    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
    #include<cstdio>
    #include<stack>
    #include<cstring>
    using namespace std;
    const int maxn = 100010;
    const int sqrN = 351;
    
    stack<int> st;
    int block[maxn],table[maxn];
    
    void peekMedian(int k){
        int sum = 0;
        int idx = 0;
        while(sum + block[idx] < k){
            sum += block[idx++];
        }
        int num = idx * sqrN;
        while(sum + table[num] < k){
            sum += table[num++];
        }
        printf("%d
    ",num);
    }
    
    void Push(int x){
        st.push(x);
        block[x/sqrN]++;
        table[x]++;
    }
    void Pop(){
        int t = st.top();
        st.pop();
        block[t/sqrN]--;
        table[t]--;
        printf("%d
    ",t);
    }
    
    int main(){
        int n,x;
        memset(block,0,sizeof(block));
        memset(table,0,sizeof(table));
        scanf("%d",&n);
        char cmd[20];
        for(int i = 0; i < n; i++){
            scanf("%s",cmd);
            if(strcmp(cmd,"Push") == 0){
                scanf("%d",&x);
                Push(x);
            }else if(strcmp(cmd,"Pop") == 0){
                if(st.empty() == true){
                    printf("Invalid
    ");
                }else{
                    Pop();
                }
            }else{
                if(st.empty() == true){
                    printf("Invalid
    ");
                }else{
                    int k = st.size();
                    if(k % 2 == 0) k = k/2;
                    else k = (k+1)/2;
                    peekMedian(k);
                }
            }
        }
        return 0;
    }
  • 相关阅读:
    采集智能电表
    未能写入输出文件“c:\WINDOWS\Microsoft.NET\Framework\.....dll”“拒绝访问。
    随笔写写jquery
    随便写写,,
    写写Ajaxpro
    C# 给程序加日志功能。
    Oracle_Database_11g_标准版_企业版__下载地址_详细列表
    通过C#发送自定义的html格式邮件
    C# 加密解密链接字符串
    获取本地 有线 正在使用的网卡信息
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/9429820.html
Copyright © 2011-2022 走看看