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;
}