杭电1040

 

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44723    Accepted Submission(s): 19105

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

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

 

Sample Output

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

 

刚开始一直超时，后来才发现数组取小了，原来这个也有关系，菜鸟真的什么都不懂啊。

冒泡排序做的

#include<stdio.h>

 2 int main()

 3 {

 4     int m,n;

 5     int i,j,k;

 6     int a[1000],temp;

 7     scanf("%d",&m);

 8     for(i=0;i<m;i++)

 9     {

10         scanf("%d",&n);

11         for(j=0;j<n;j++)

12         scanf("%d",&a[j]);

13         for(j=0;j<n-1;j++)//冒泡排序； 14         {

15             

16             for(k=j+1;k<n;k++)

17             {

18                 if(a[j]>=a[k])

19                 {

20                     temp=a[j];

21                     a[j]=a[k];

22                     a[k]=temp;

23                 }

24             }

25         }

26         printf("%d",a[0]);

27         for(j=1;j<n;j++)

28         printf(" %d",a[j]);

29         printf("\n");

30         

31     }

32     

33 }

 插入排序做的

#include<stdio.h>

int main()

{

    int m,n;

    int i,j,k;

    int a[1000],temp,out,in;

    scanf("%d",&m);

    for(i=0;i<m;i++)

    {

        scanf("%d",&n);

        for(j=0;j<n;j++)

        scanf("%d",&a[j]);

        for(out=1;out<n;out++)

        {

            temp=a[out];

            in=out;

            while(in>0&&a[in-1]>=temp)

            {

                a[in]=a[in-1];

                in--;

            }

            a[in]=temp;

        }

        printf("%d",a[0]);

        for(j=1;j<n;j++)

        printf(" %d",a[j]);

        printf("\n");

        

    }

    

}