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. 

3
1 2 3
4 
4 3 2 1 

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。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 int a[1010];
 5 int main(){
 6     int n;
 7     while(scanf("%d",&n)!=EOF){
 8         for(int i = 0; i < n; i ++){
 9             //scanf("%d",a[i]);
10             cin>>a[i];
11         }
12         ll ans = 0;
13         for(int i =1; i < n; i ++){ //13到17行为冒泡排序法
14             for(int j = 0; j <n-1; j ++){
15                 if(a[j]>a[j+1]){
16                     ans++;
17                     swap(a[j],a[j+1]);
18                 }
19             }
20         }
21         cout << ans << endl;
22         //printf("")
23     }
24     return 0;
25 }