链接：click here

题意：

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 T number of test cases.(<0T<1000) 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

2
3
1 2 3
4
4 3 2 1

样例输出

0
6

思路：就是求一组数据的逆序数，树状数组求法，不解释：

代码：

#include 
    
     
#include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include 
              
                using namespace std; #define lowbit(a) a&-a #define Max(a,b) a>b?a:b #define Min(a,b) a>b?b:a #define mem(a,b) memset(a,b,sizeof(a)) int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}}; const double eps = 1e-6; const double Pi = acos(-1.0); static const int inf= ~0U>>2; static const int maxn =110; int h[1000001],w[100],Map[200]; int m[10001], N; void Add(int i) { while(i<=1001) { m[i]++; i+=lowbit(i); } } int Sum(int i) { int res=0; while(i>0) { res+=m[i]; i-=lowbit(i); } return res; } int main() { int T,temp; scanf("%d",&T); while(T--) { scanf("%d",&N); memset(m,0,sizeof(m)); int ans=0; for(int i=1; i<=N; i++) { scanf("%d",&temp); Add(temp); ans+=(i-Sum(temp)); } printf("%d\n",ans); } return 0; } 
              
             
            
           
          
         
        
       
      
     
    

When you want to give up, think of why you persist until now！