HDOJ 1040 As Easy As A+B(qsort)
2015-07-21 11:17
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As Easy As A+B
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 45864 Accepted Submission(s): 19588
[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
[align=left]Recommend[/align]
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#include<stdio.h> #include<stdlib.h> #include<string.h> int a[1001]; int cmp(const void *a,const void *b){ return *(int *)a-*(int *)b;//升序 } int main(){ int n,i,m; scanf("%d",&n); while(n--){ scanf("%d",&m); for(i=0;i<m;i++) scanf("%d",&a[i]); qsort(a,m,sizeof(a[0]),cmp); for(i=0;i<m-1;i++) printf("%d ",a[i]); printf("%d\n",a[m-1]); } return 0; }
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