zoukankan      html  css  js  c++  java
  • Codeforces Round #605 (Div. 3) A. Three Friends(贪心)

    链接:

    https://codeforces.com/contest/1272/problem/A

    题意:

    outputstandard output
    Three friends are going to meet each other. Initially, the first friend stays at the position x=a, the second friend stays at the position x=b and the third friend stays at the position x=c on the coordinate axis Ox.

    In one minute each friend independently from other friends can change the position x by 1 to the left or by 1 to the right (i.e. set x:=x−1 or x:=x+1) or even don't change it.

    Let's introduce the total pairwise distance — the sum of distances between each pair of friends. Let a′, b′ and c′ be the final positions of the first, the second and the third friend, correspondingly. Then the total pairwise distance is |a′−b′|+|a′−c′|+|b′−c′|, where |x| is the absolute value of x.

    Friends are interested in the minimum total pairwise distance they can reach if they will move optimally. Each friend will move no more than once. So, more formally, they want to know the minimum total pairwise distance they can reach after one minute.

    You have to answer q independent test cases.

    思路:

    中间的值影响不大,手动判断条件。

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long LL;
     
    int main()
    {
        int t;
        cin >> t;
        while(t--)
        {
            int a[4];
            for (int i = 1;i <= 3;i++)
                cin >> a[i];
            sort(a+1, a+4);
            if (a[1] == a[2])
            {
                if (a[3] > a[2])
                    a[3]--;
                if (a[1] < a[3])
                    a[1]++, a[2]++;
            }
            else if (a[2] == a[3])
            {
                if (a[1] < a[2])
                    a[1]++;
                if (a[2] > a[1])
                    a[2]--, a[3]--;
            }
            else
            {
                a[1]++;
                a[3]--;
            }
            int ans = a[2]-a[1]+a[3]-a[1]+a[3]-a[2];
            cout << ans << endl;
        }
     
        return 0;
    }
    
  • 相关阅读:
    HDOJ 2577 How To Type
    HDOJ 1171 Big Event in HDU
    HDOJ 2159 FATE
    HDOJ 1176 免费馅饼
    POJ 1014 Dividing
    HDOJ 2844 Coins
    可以设置DefaultButton的TextBox控件
    setTimeout和setInterval的使用
    C# 调用ExchangeWebservice的相关代码
    实现IConfigurationSectionHandler接口来编写自定义配置
  • 原文地址:https://www.cnblogs.com/YDDDD/p/12046656.html
Copyright © 2011-2022 走看看