UVA1588Kickdown

一家世界领先的汽车公司的研究实验室接到任务,需要创建一种特殊的传动机制,该机制能够实现高效的换挡操作。经过数月的研究,工程师们发现最有效的解决方案需要使用具有不均匀分布齿和空腔的特殊齿轮,并已经计算出了齿轮的最佳轮廓。为了验证这一发现,他们计划进行实验,需要从昂贵的合金材料中切割出两个不规则但可以相互配合的齿条部分。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A research laboratory of a world-leading automobile company has received an order to create a specialtransmission mechanism, which allows for incredibly efficient kickdown — an operation of switching tolower gear. After several months of research engineers found that the most efficient solution requiresspecial gears with teeth and cavities placed non-uniformly. They calculated the optimal flanks of thegears. Now they want to perform some experiments to prove their findings.

The first phase of the experiment is done with planar toothed sections, not round-shaped gears. Asection of length n consists of n units. The unit is either a cavity of height h or a tooth of height 2h.Two sections are required for the experiment: one to emulate master gear (with teeth at the bottom)and one for the driven gear (with teeth at the top).

There is a long stripe of width 3h in the laboratory and its length is enough for cutting two engagedsections together. The sections are irregular but they may still be put together if shifted along eachother.

The stripe is made of an expensive alloy, so the engineers want to use as little of it as possible. Youneed to find the minimal length of the stripe which is enough for cutting both sections simultaneously.

Input

The input file contains several test cases, each of them as described below.

There are two lines in the input, each contains a string to describe a section. The first line describesmaster section (teeth at the bottom) and the second line describes driven section (teeth at the top).Each character in a string represents one section unit — 1 for a cavity and 2 for a tooth. The sectionscan not be flipped or rotated.

Each string is non-empty and its length does not exceed 100.

Output

For each test case, write to the output a line containing a single integer number — the minimal lengthof the stripe required to cut off given sections.

Sample Input

2112112112

2212112

12121212

21212121

2211221122

21212

Sample Output

10

8

15

代码:

#include <stdio.h>  
#include <cstdlib>  
#include <string.h>  
#include <iostream>  
using namespace std; 
int f(char s[], char s1[])
{  
	int len,len1,i,j,flag;
	len=strlen(s);
	len1=strlen(s1);
	for(i=0;i<len;i++)
	{  
		flag=1;
		for (j=i;j<len&&j<i+len1;j++)
			if (s[j]=='2'&&s1[j-i]=='2')
				flag=0;
        if(flag)
			return max(len,i+len1);
    }
	return len+len1;
}  
int main()
{
	int len;
	char s[101],s1[101];
	while(scanf("%s%s",s,s1)!=EOF)
	{
		len = min(f(s, s1),f(s1, s));
		printf("%d\n",len);
	}
	return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

kunsir_

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值