NYOJ - sort it

Sort it

时间限制：1000 ms  |           内存限制：65535 KB

难度：2

描述

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.

输入

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.

输出

For each case, output the minimum times need to sort it in ascending order on a single line.

样例输入

3
1 2 3
4 
4 3 2 1 

样例输出

0
6

#include <stdio.h>

int sort_count(int a[], int n)      // 冒泡排序 
{
	int count = 0;
	for(int i = 0; i < n-1; i++)
	{
		for(int j = 0; j < n-1-i; j++)
		{
			if(a[j] > a[j+1])
			{
				int t = a[j];
				a[j] = a[j+1];
				a[j+1] = t;
				count++;
			}
		}
	}
	return count;
}
int main()
{
	int n,a[1000];
	while(scanf("%d",&n) != EOF)
	{
		for(int i = 0; i < n; i++)
		{
			scanf("%d",&a[i]);
		}
		printf("%d\n",sort_count(a,n));
	}
}