杭电Sort it 2689树状数组

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3510    Accepted Submission(s): 2546


Problem Description
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.
 

Input
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.
 

Output
For 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
//虽然是水题,但是对于初学者来说很想通,自己好好推理推理
#include <iostream> #include <cstring> using namespace std; const int maxn=1010; int C[maxn],n; int lowbit(int x) {     return x&(-x); } void update(int i) {     while(i<=n)     {         C[i]+=1;         i+=lowbit(i);     } } int query(int i) {     int sum=0;     while(i>0)     {         sum+=C[i];         i-=lowbit(i);     }     return sum; } int main() {     while(cin>>n)     {         int sum=0;         memset(C,0,sizeof(C));         for(int i=1;i<=n;i++)         {             int a;            cin>>a;             update(a);             sum+=i-query(a);//得出比i大的的元素个数,即要交换的次数         }     cout<<sum<<endl;     }     return 0; }
 

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