The second chapter on the Computer Experiment report

1. Practical Topics:

Median number of two ordered sequences

2. Title Description:

Knowing that there are two equal-length non-descending sequence S1, S2, the design function S1 and S2 The median of the set. Ordered sequenceA?0??,A?1??,?,A?N?1?? The median digit of theA?(N?1)/2?? Value, that is, the first? (n+1) /2 number ( a ? 0 ?? is the 1th number).

Input format:

The input is divided into three lines. The first line gives the public length of the sequence N (0<n≤100000), followed by each line entering a sequence of information, i.e. n non-descending integers. The number is separated by a space.

Output format:

Outputs the median of the sequence of two input sequences in a row.

Input Sample 1:

51 3 5 7 92 3 4 5 6

Output Example 1:

4

Input Sample 2:

6-100 -10 1 1 1 1-50 0 2 3 4 5

Output Example 2:

1

3. Algorithm Description:

Using the idea of dichotomy, first to find the median of each sequence a[m1],b[m2], compare the size of a[m1],b[m2], if A[M1]==B[M2], you can determine the median of the merge sequence is a[m1], because A[M1] is greater than a[m1] in front of the element, less than a[ M1] The element that follows, B[M2] is greater than b[m2], and the element after b[m2], and a[m1] is equal to b[m2], you can determine that A[M1] is in the middle of the merge sequence. If A[M1]<B[M2], in two cases: if each sequence is an odd number, then Start1=m1,end2=mid2 (before the big take, after the small take), if it is an even number, then start1=m1+1,end2=m2. If A[M1]>B[M2], also divided into two cases, the same is large before, after the small take, odd: end1=m1,start2=m2; even: end1=m1,end2=m2+1. As long as the start1<end1&&start3<end2, it has been approximated by the dichotomy, until only two numbers are left, the smaller number is the median.

Code implementation:

#include <iostream>
using namespace std;
int m_search (int a[], int b[], int n)
{
int start1 = 0, End1 = n-1, m1, Start2 = 0, End2 = n-1, m2;

while (start1! = End1 | | Start2! = end2)
{
M1 = (Start1 + end1)/2;
M2 = (Start2 + end2)/2;
if (a[m1] = = b[m2])
return A[M1];

if (a[m1]<b[m2])
{
if ((Start1 + end1)% 2 = = 0)
{
Start1 = m1;
End2 = m2;
}
Else
{
Start1 = m1 + 1;
End2 = m2;
}
}
Else
{
if ((Start2 + end2)% 2 = = 0)
{
END1 = m1;
Start2 = m2;
}
Else
{
END1 = m1;
Start2 = m2 + 1;
}
}
}
return A[start1]<b[start2]? A[START1]: B[start2];
}
int main () {
int a[100001],b[100001];
int n;
Cin>>n;
for (int i=0;i<n;i++)
{
cin >>a[i];
}
for (int i=0;i<n;i++)
{
Cin>>b[i];
}
Cout<<m_search (A, B, N);
return 0;
}

4, time complexity and spatial complexity analysis:

Time complexity: The time complexity of the dichotomy is O (logn)

Space complexity: O (1)

5, Experience:

When I first saw the problem, I thought I was going to start by putting two sequences in order and then looking for the median of the sorted array, that is, the fast-track, and then I found that the time complexity of this approach was O (NLOGN). Then think of just learn the dichotomy, but in the case of odd and even card for a long time, and finally slowly try to come to the conclusion.

The second chapter on the Computer Experiment report