codeforce#377C. Sanatorium

最新推荐文章于 2024-05-04 16:55:46 发布

Rain722

最新推荐文章于 2024-05-04 16:55:46 发布

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本文探讨了一道关于敬老院用餐记录的问题，通过分析输入的早餐、午餐和晚餐次数，帮助忘记度假细节的人计算出可能错过的最少餐数。文章提供了两种解题思路及代码实现。

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Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!

Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.

Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.

According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.

Input

The only line contains three integers b, d and s (0 ≤ b, d, s ≤ 1018, b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.

Output

Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.

Examples

input
3 2 1

output
1


input
1 0 0

output
0


input
1 1 1

output
0


input
1000000000000000000 0 1000000000000000000

output
999999999999999999

Note

In the first sample, Vasiliy could have missed one supper, for example, in case he have arrived before breakfast, have been in the sanatorium for two days (including the day of arrival) and then have left after breakfast on the third day.

In the second sample, Vasiliy could have arrived before breakfast, have had it, and immediately have left the sanatorium, not missing any meal.

In the third sample, Vasiliy could have been in the sanatorium for one day, not missing any meal.

题意：来敬老院的这些天中，除了最后一天与第一天，其中的日子里是整天都呆着的！！！！

思路：分类的话，就是第一天的要么吃中餐和晚餐，要么吃晚餐，要么第一天都吃，最后一天要么吃早餐，要么吃早餐和中餐，要么全吃。

第一天和最后一天的情况如下:

3 3

1 3

3 1

1 1

2 1

1 2

2 3

3 2

2 2

其实可以看到，如果把第一天和最后一天吃的挪到同一天上来，总归有一餐漏吃或有一餐多吃，其他的几餐都是相等的，然而只有一餐漏吃或者多吃是被允许的，可以思考下！

智障的孩子，智障的AC：

#include<bits/stdc++.h>
using namespace std;
long long int a[4], b[4], d[4];
int main()
{
    while(cin >> a[0] >> a[1] >> a[2])
    {
        d[0] = a[0]; d[1] = a[1]; d[2] = a[2];
        sort(d, d+3);
        if(d[0] == 0 && d[1] == 0 && d[2] == 1)
        {
            cout << 0 << endl;
            continue;
        }
        if(a[0] == 0 && a[2] == 0 && a[1])
        {
            cout << 2*a[1] - 2 << endl;
            continue;
        }
        long long int ans = 1e20, tem;
        for(int i = 0; i <= 7; i++)
        {
            b[0] = a[0];b[1] = a[1]; b[2] = a[2];
            if(i == 0)      // 0 0
            {
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
            if(i == 1)      //1 0
            {
                if(b[0] > 0)
                    b[0]--;
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
            if(i == 2)      // 0  1
            {
                if(b[2] > 0)
                    b[2]--;
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
            if(i == 3)      // 1 1
            {
                if(b[0] > 0 && b[2] > 0)
                    b[0]--, b[2]--;
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
            if(i == 4)      //2 1
            {
                if(b[0] >= 1 &&b[1] >= 1 && b[2] >= 1)
                    b[0]--, b[1]--, b[2]--;
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
            if(i == 5)      //2 0
            {
                if(b[1] >= 1 && b[2] >=1)
                    b[1]--, b[2]--;
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
            if(i == 6)
            {
                if(b[0] >= 1 && b[1] >=1)
                    b[0]--, b[1]--;;
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
            if(i == 7)
            {
                if(b[2] >=1 && b[0] >= 1 && b[1] >= 2)
                    b[1] -= 2, b[0]--, b[2]--;
                sort(b, b + 3);
                tem = 2 * b[2] - b[1] - b[0];
                ans = min(tem, ans);
            }
        }
        cout << ans << endl;
    }
    return 0;
}

笨的孩子只有慢慢学：

#include<bits/stdc++.h>
using namespace std;
long long a[4];
int main()
{
    long long dif1, dif2, ans;
    while(cin >> a[1] >> a[2] >> a[3])
    {
        ans = 0;
        sort(a+1, a+3+1);
        a[3]--;
        dif1 = a[3] - a[2];
        if(dif1 > 0)
            ans += dif1;
        dif2 = a[3] - a[1];
        if(dif2 > 0)
            ans += dif2;
        cout << ans << endl;
    }
    return 0;
}