杭电acm 1040 As Easy As A+B
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杭电acm 1040
As Easy As A+B
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22439 Accepted Submission(s): 9357
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!
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.
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
我的accept代码:
#include <iostream>
using namespace std;
int main()
{
int a[1001],num;
int i,temp;
cin>>num;
while(num--)
{
memset(a,0,sizeof(a));
i=0;
cin>>i;
for(int q=0;q<i;q++)
cin>>a[q];
for(int j=0;j<i-1;j++)//冒泡排序法,最大的在最后
for(int k=0;k<i-1-j;k++)
{
if(a[k]>a[k+1])
{
temp = a[k+1];
a[k+1] = a[k];
a[k] = temp;
}
}
for(int z=0;z<i-1;z++)
cout<<a[z]<<" ";//注意别把a[i-1]放在这里输出,不然就有最后一项有" ",这样会错误
cout<<a[i-1]<<endl;
}
return 0;
}
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