HDU 1040 As Easy As A+B

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52312    Accepted Submission(s): 22494

Problem Description

These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.
Give you some integers, your task is to sort these number ascending (升序).
You should know how easy the problem is now!
Good luck!

 

Input

Input contains multiple test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains an integer N (1<=N<=1000 the number of integers to be sorted) and then N integers follow in the same line. 
It is guarantied that all integers are in the range of 32-int.

 

Output

For each case, print the sorting result, and one line one case.

 

Sample Input

23 2 1 39 1 4 7 2 5 8 3 6 9

 

Sample Output

1 2 31 2 3 4 5 6 7 8 9

 

Author

lcy

 

题意看上去是一道水题。

以此题来复习排序算法

堆排序

#include <stdio.h>#include <algorithm>using namespace std;const int MAXN=1005;int num[MAXN];void percDown(int a[],int i,int n){        int child;        int temp;        for(temp=a[i];2*i+1<n;i=child)        {                child=2*i+1;                if(child!=n-1&&a[child+1]>a[child])                        child++;                if(temp<a[child])                        a[i]=a[child];                else                        break;        }        a[i]=temp;}void heapSort(int a[],int n){        for(int i=n/2;i>=0;i--)                percDown(a,i,n);        for(int i=n-1;i>0;i--)        {                swap(a[0],a[i]);                percDown(a,0,i);        }}int main(){        int t;        scanf("%d",&t);        while(t--)        {                int n;                scanf("%d",&n);                for(int i=0;i<n;i++)                        scanf("%d",&num[i]);                heapSort(num,n);                for(int i=0;i<n;i++)                        printf("%d%c",num[i],i==n-1?'\n':' ');        }        return 0;}

归并排序

分治的思想

#include <stdio.h>#include <stdlib.h>const int MAXN=1005;int num[MAXN];void Merge(int a[],int temp[],int Lpos,int Rpos,int RightEnd){        int LeftEnd=Rpos-1;        int nums=RightEnd-Lpos+1;        int tempPos=Lpos;        while(Lpos<=LeftEnd&&Rpos<=RightEnd)                if(a[Lpos]<=a[Rpos])                        temp[tempPos++]=a[Lpos++];                else                        temp[tempPos++]=a[Rpos++];        while(Lpos<=LeftEnd)                temp[tempPos++]=a[Lpos++];        while(Rpos<=RightEnd)                temp[tempPos++]=a[Rpos++];        for(int i=0;i<nums;i++,RightEnd--)                a[RightEnd]=temp[RightEnd];}void Msort(int a[],int temp[],int left,int right){        if(left<right)        {                int center=(left+right)/2;                Msort(a,temp,left,center);                Msort(a,temp,center+1,right);                Merge(a,temp,left,center+1,right);        }}void mergeSort(int a[],int n){        int *temp=(int *)malloc(n*sizeof(int));        if(temp!=NULL)        {                Msort(a,temp,0,n-1);                free(temp);        }}int main(){        int t;        scanf("%d",&t);        while(t--)        {                int n;                scanf("%d",&n);                for(int i=0;i<n;i++)                        scanf("%d",&num[i]);                mergeSort(num,n);                for(int i=0;i<n;i++)                        printf("%d%c",num[i],i==n-1?'\n':' ');        }        return 0;}