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:
    损失函数 代价函数 评分函数 目标函数
    python目录索引
    机器学习/深度学习资料合集
    Git笔记
    目标检测中的正负样本分配
    map计算
    nms
    08shell脚本
    07makefile文件
    05-STL
  • 原文地址:https://www.cnblogs.com/YDDDD/p/12046656.html
Copyright © 2011-2022 走看看