NYOJ - sort it-优快云博客

本文介绍了一个简单的排序计数算法，该算法通过冒泡排序的方式计算将一组乱序的整数序列排序所需的最少交换次数。文章包含完整的C语言实现代码，并提供了一个具体的示例输入输出以展示算法的工作原理。

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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

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));
	}
}