zoukankan      html  css  js  c++  java
  • CF Polycarpus' Dice (数学)

    Polycarpus' Dice
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Polycarp has n dice d1, d2, ..., dn. The i-th dice shows numbers from 1 to di. Polycarp rolled all the dice and the sum of numbers they showed is A. Agrippina didn't see which dice showed what number, she knows only the sum A and the values d1, d2, ..., dn. However, she finds it enough to make a series of statements of the following type: dice i couldn't show number r. For example, if Polycarp had two six-faced dice and the total sum is A = 11, then Agrippina can state that each of the two dice couldn't show a value less than five (otherwise, the remaining dice must have a value of at least seven, which is impossible).

    For each dice find the number of values for which it can be guaranteed that the dice couldn't show these values if the sum of the shown values is A.

    Input

    The first line contains two integers n, A (1 ≤ n ≤ 2·105, n ≤ A ≤ s) — the number of dice and the sum of shown values where s = d1 + d2 + ... + dn.

    The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 106), where di is the maximum value that the i-th dice can show.

    Output

    Print n integers b1, b2, ..., bn, where bi is the number of values for which it is guaranteed that the i-th dice couldn't show them.

    Sample test(s)
    input
    2 8
    4 4
    output
    3 3 
    input
    1 3
    5
    output
    4 
    input
    2 3
    2 3
    output
    0 1 



    当前骰子最小值 + 剩余筛子最大值 >= A
    当前骰子最大值 + 剩余筛子最小值 <= A
    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <cstdio>
    #include <string>
    #include <algorithm>
    #include <cctype>
    #include <queue>
    #include <map>
    using    namespace    std;
    
    int    main(void)
    {
        int        n;
        int        s[200005];
        long    long    A;
        long    long    sum = 0;
        long    long    ans = 0;
    
        cin >> n >> A;
        for(int i = 0;i < n;i ++)
        {
            cin >> s[i];
            sum += s[i];
        }
        for(int i = 0;i < n;i ++)
        {
            long    long    m_min = A - sum + s[i];
            long    long    m_max = A - (n - 1);
            if(m_max > s[i])
                m_max = s[i];
            ans = s[i] - m_max + (m_min - 1 > 0 ? m_min - 1: 0);
            cout << ans;
            if(i != n - 1)
                cout << ' ';
        }
        cout << endl;
    
        return    0;
    }
  • 相关阅读:
    quick cocos2dx 播放序列帧动画的实现
    PS学习笔记
    python自学笔记(十一)关于函数及书写格式
    python自学笔记(十)语句与数据结构应用
    python自学笔记(九)python练习题
    python自学笔记(八)python语句
    python自学笔记(七)排序与多级排序
    python自学笔记(六)二进制与位移
    python自学笔记(五)python文本操作
    python自学笔记(四)python基本数据类型之元组、集合、字典
  • 原文地址:https://www.cnblogs.com/xz816111/p/4438962.html
Copyright © 2011-2022 走看看