CodeFoeces732C 枚举

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。

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;
}
传送带损坏与对象检测数据集 一、基础信息 • 数据集名称:传送带损坏与对象检测数据集 • 图片数量: 训练集:645张图片 验证集:185张图片 测试集:92张图片 总计:922张工业监控图片 • 训练集:645张图片 • 验证集:185张图片 • 测试集:92张图片 • 总计:922张工业监控图片 • 分类类别: Hole(孔洞):传送带表面的孔洞损坏。 Human(人类):工作区域中的人类,用于安全监控。 Other Objects(其他对象):非预期对象,可能引起故障。 Puncture(刺穿):传送带被刺穿的损坏。 Roller(滚筒):传送带滚筒部件。 Tear(撕裂):传送带撕裂损坏。 impact damage(冲击损坏):由于冲击导致的损坏。 patch work(修补工作):已修补的区域。 • Hole(孔洞):传送带表面的孔洞损坏。 • Human(人类):工作区域中的人类,用于安全监控。 • Other Objects(其他对象):非预期对象,可能引起故障。 • Puncture(刺穿):传送带被刺穿的损坏。 • Roller(滚筒):传送带滚筒部件。 • Tear(撕裂):传送带撕裂损坏。 • impact damage(冲击损坏):由于冲击导致的损坏。 • patch work(修补工作):已修补的区域。 • 标注格式:YOLO格式,包含边界框和类别标签,适用于目标检测任务。 • 数据格式:图像数据来源于工业监控系统,适用于计算机视觉分析。 二、适用场景 • 工业自动化检测系统开发:用于构建自动检测传送带损坏和异物的AI模型,实现实时监控和预防性维护,减少停机时间。 • 安全监控应用:识别人类和其他对象,提升工业环境的安全性,避免事故和人员伤害。 • 学术研究与创新:支持计算机视觉在制造业、物流和自动化领域的应用研究,促进AI技术与工业实践的融合。 • 教育与培训:可用于培训AI模型或作为工业工程和自动化教育的案例数据,帮助学习者理解实际应用场景。 三、数据集优势 • 多样化的类别覆盖:包含8个关键类别,涵盖多种损坏类型和对象,确保模型能够处理各种实际工业场景,提升泛化能力。 • 精准的标注质量:采用YOLO格式,边界框标注准确,由专业标注人员完成,保证数据可靠性和模型训练效果。 • 强大的任务适配性:兼容主流深度学习框架(如YOLO、TensorFlow、PyTorch),可直接用于目标检测任务,并支持扩展至其他视觉任务需求。 • 突出的工业价值:专注于工业传送带系统的实际需求,帮助提升生产效率、降低维护成本,并增强工作场所安全,具有较高的实际应用价值。
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