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): 46460    Accepted Submission(s): 19869


[align=left]Problem Description[/align]
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!

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each case, print the sorting result, and one line one case.

 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

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

 

[align=left]Author[/align]
lcy
 

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#include<stdio.h>
#include<stdlib.h>
int cmp(const void*a,const void*b){
return(*(int*)a-*(int*)b);
}
int main(){
int n;
scanf("%d",&n);
while(n--){
int N,a[1005],i;
scanf("%d",&N);
for(i=0;i<N;i++)
scanf("%d",&a[i]);
qsort(a,N,sizeof(a[0]),cmp);
for(i=0;i<N;i++){
printf("%d",a[i]);
if(i!=N-1)
printf(" ");
}//注意：最后一个元素后无空格直接换行
printf("\n");
}
return 0;
}