C. Sanatorium
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
没啥好说的疯狂枚举
题干难懂恶心人
#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<string>
#include<memory.h>
#include<stack>
using namespace std;
int main()
{
long long q,w,e;
cin>>q>>w>>e;
long long sum=2111111111111111111;
long long qq=q,ww=w,ee=e;
qq=0,ww=w-q+1,ee=e-q+1;
long long zanshi=0;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
qq=0,ww=w-q,ee=e-q+1;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
qq=0,ww=w-q+1,ee=e-q;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
qq=0,ww=w-q,ee=e-q;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ww=0,qq=q-w+1,ee=e-w+1;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ww=0,qq=q-w+1,ee=e-w;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ww=0,qq=q-w,ee=e-w+1;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ww=0,qq=q-w,ee=e-w;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ee=0,ww=w-e+1,qq=q-e+1;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ee=0,ww=w-e,qq=q-e+1;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ee=0,ww=w-e+1,qq=q-e;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
ee=0,ww=w-e,qq=q-e;
if(qq<=0&&ww<=0&&ee<=0)
{
zanshi=qq+ww+ee;
zanshi=-zanshi;
sum=min(sum,zanshi);
zanshi=0;
}
cout<<sum<<endl;
return 0;
}

本文介绍了一种通过解析记录在卡片上的餐食次数来确定某人在疗养院期间可能错过的最少餐食数量的方法。利用枚举策略,文章提供了一个算法实现的例子,展示了如何根据早餐、晚餐和夜宵的数量来计算缺失的餐食数。
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