Sort it

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. 

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

Sample Input

3
1 2 3
4 
4 3 2 1 

Sample Output

0
6
大概意思：给有序数组里相邻的数字进行排序，算出排序时需要进行的最小步骤数。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int a[1010];
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i = 0; i < n; i ++){
            //scanf("%d",a[i]);
            cin>>a[i];
        }
        ll ans = 0;
        for(int i =1; i < n; i ++){ //13到17行为冒泡排序法
            for(int j = 0; j <n-1; j ++){
                if(a[j]>a[j+1]){
                    ans++;
                    swap(a[j],a[j+1]);
                }
            }
        }
        cout << ans << endl;
        //printf("")
    }
    return 0;
}