Running Median

Time Limit: 1 Sec??Memory Limit: 256 MB

題目連接

　　http://acm.hdu.edu.cn/showproblem.php?pid=3282

Description

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.

Input

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.

Output

For each data set the first line of output contains the data set number, a single space and the number of medians output (which should be one-half the number of input values plus one). The output medians will be on the following lines, 10 per line separated by a single space. The last line may have less than 10 elements, but at least 1 element. There should be no blank lines in the output.

Sample Input

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

Sample Output

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

HINT

題意

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??給你n個數，每次插入一個數，當插入數的數量為奇數的時候，我們就輸出中位數

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題解：

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這道題要求動態求中位數，但是數據范圍太小了，于是我們直接暴力搞就好了

事先排序，然后每次都掃一遍就行了……

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代碼：

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//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 100005 #define mod 10007 #define eps 1e-9 //const int inf=0x7fffffff; //無限大 const int inf=0x3f3f3f3f; /**/ //************************************************************************************** inline ll read() {int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f; } struct node {int x,y; }; node a[maxn]; bool cmp(node b,node c) {return b.x<c.x; } int main() {int t=read();for(int cas=1;cas<=t;cas++){vector<int> ans;int n=read(),m=read();for(int i=0;i<m;i++)a[i].x=read(),a[i].y=i+1;sort(a,a+m,cmp);for(int i=0;i<m;i+=2){int flag=0;for(int j=0;j<m;j++){if(a[j].y<=i+1)flag++;if(flag==(i+2)/2){ans.push_back(a[j].x);break;}}}printf("%d %d",cas,ans.size());for(int i=0;i<ans.size();i++){if((i)%10==0)printf("\n");if(i%10==0)printf("%d",ans[i]);elseprintf(" %d",ans[i]);}printf("\n");} }

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轉載于:https://www.cnblogs.com/qscqesze/p/4386718.html

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