hdu 2689 sort it

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4487    Accepted Submission(s): 3119

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

31 2 34 4 3 2 1 

 

Sample Output

0
6
哪个数进入序列 就把他置成1 不是第几个输进来把第几个置成1 而是根据他的数值确定位置 所以建立的树状数组是排好序的 。
答案就是每个数前面比她大的个数和 
这个数是i-query(n) i代表是第几个数 n 是这个数的数值。
#include <iostream>#include<stdio.h>#include<string.h>using namespace std;const int  maxn=1005;int lowbit(int i){    return i&(-i);}int a[maxn];int n;int update(int i,int b){    while(i<=n)    {        a[i]+=b;        i+=lowbit(i);    }}int query(int i){    int sum=0;    while(i>0)    {        sum+=a[i];        i=i-lowbit(i);    }    return sum;}int main(){    while(cin>>n)    {        int ans=0,x;        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {            scanf("%d",&x);            update(x,1);            ans+=i-query(x);        }        cout<<ans<<endl;    }    return 0;}