hdu 2689 Sort it（树状数组）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2689点击打开链接

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4693    Accepted Submission(s): 3261


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

 
暴力可以过的题 用树状数组求逆序对写了下

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include<algorithm>
#include <math.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <vector>
using namespace std;
int tree[1000000];
int lowbit (int x)
{
return x&-x;
}
int getsum(int x)
{
int sum=0;
while(x>0)
{
sum+=tree[x];
x-=lowbit(x);
}
return sum;
}
void add(int x,int num)
{
while(x<1000000)
{
tree[x]+=num;
x+=lowbit(x);
}
}
int main()
{

b956
int n=0;int s=0;
while(~scanf("%d",&n))
{
s=0;
memset(tree,0,sizeof(tree));
for(int i=0;i<n;i++)
{
int nn=0;
scanf("%d",&nn);
add(nn,1);
s+=(getsum(n)-getsum(nn));
}
cout << s << endl;
}

}