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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> CPP 1373 Easy as A+B(冒泡排序)

CPP 1373 Easy as A+B(冒泡排序)

編輯:C++入門知識

CPP 1373 Easy as A+B(冒泡排序)


 

題面:

 

Easy as A+B
Time Limit:1000MS Memory Limit:32768K

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. 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 32bit-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

Source:

lcy

題解:

冒泡排序一下,套了下別人寫的,居然是錯的,以後還是自己寫吧。

 

代碼:

 

#include 
int main() 
{ 
  int temp,t,n; 
  int store[1005]; 
  scanf(%d,&t);
  while(t--)
  {
    scanf(%d,&n);
    for(int i=0;i store[j])
                    {
                        temp = store[i];
                        store[i] = store[j];
                        store[j] = temp;
                    }
                }
            }
    printf(%d,store[0]);
    for(int i=1;i

 

 

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