2014
03-13

# Running Median

For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far.

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer M, (1 ≤ M ≤ 9999), giving the total number of signed integers to be processed.
The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space.
The last line in the dataset may contain less than 10 values.

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer M, (1 ≤ M ≤ 9999), giving the total number of signed integers to be processed.
The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space.
The last line in the dataset may contain less than 10 values.

3
1 9
1 2 3 4 5 6 7 8 9
2 9
9 8 7 6 5 4 3 2 1
3 23
23 41 13 22 -3 24 -31 -11 -8 -7
3 5 103 211 -311 -45 -67 -73 -81 -99
-33 24 56

1 5
1 2 3 4 5
2 5
9 8 7 6 5
3 12
23 23 22 22 13 3 5 5 3 -3
-7 -3

//hdu 3282 Running Median
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;

void run()
{
int n,m;
scanf("%d%d",&n,&m);
int i,num[10001];
int ans[10001]={0},
int u=2;

printf("%d %d/n",n,(m+1)/2);
scanf("%d",&num[1]);
ans[1]=num[1];
for(i=2;i<=m;i++)
{
int tmp,j,k;
scanf("%d",&tmp);
if(tmp<=num[1]) //比第一个小，则后面的往后移
{
for(k=i-1;k>=1;k--)
{
num[k+1]=num[k];
}
num[1]=tmp;
}
else if(tmp>=num[i-1]) //比最后一个大
{num[i]=tmp;}
else
{
bool f=false;
for(j=i-1;j>=1;j--)
{
if(tmp>=num[j-1] && tmp<=num[j]) //找要插入的位置位置
{
f=true;
int k;
{
for(k=i-1;k>=j;k--)
{
num[k+1]=num[k];
}
num[j]=tmp;
}
}
if(f==true) break;
}
}
if(i%2==1) //记录奇数位置
{
ans[u++]=num[(i+1)/2];
}
}
for(i=1;i<u;i++)
{
if(i%10==0 || i==u-1) printf("%d/n",ans[i]);
else printf("%d ",ans[i]);
}
}
int main()
{
int total;
cin>>total;
for(int now=1;now<=total;now++)  run();

return 0;
}