PAT甲级1008部分正确情况分析

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题目:

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input Specification:
Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.

Output Specification:
For each test case, print the total time on a single line.

Sample Input:

3 2 3 1

Sample Output:

41

正确结果代码参考

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#include<iostream>
using namespace std;
const int maxn=110;
int main()
{
int n,a[maxn];
cin>>n;
for (int i = 0; i < n; ++i)
{
cin>>a[i];
}
int ans=0;
if(a[0]>0)
ans+=6*a[0];
ans+=5*n;
for (int j = 1; j < n; ++j)
{
if(a[j-1]<a[j])
{
ans+=6*(a[j]-a[j-1]);
}
if(a[j-1]>a[j])
{
ans+=4*(a[j-1]-a[j]);
}
}
cout<<ans;
}

以下为结果部分正确的代码

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#include<iostream>
using namespace std;
const int maxn=110;
int main()
{
int n,a[maxn];
cin>>n;
for (int i = 0; i < n; ++i)
{
cin>>a[i];
}
int P=0,ans=0;
for (int j = 0; j < n; ++j)
{
if(P<a[j])
{
ans+=6*(a[j]-P)+5;
}
if(P>a[j])
{
ans+=4*(P-a[j])+5;
}
P=a[j];
}
cout<<ans;
}

主要的改动点在for循环中每次停留时间+5秒的操作,以及设立当前电梯所处位置的参考变量,没有这两个反而能通过。

文章目录
  1. 题目:
  2. 正确结果代码参考
  3. 以下为结果部分正确的代码
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