HDU 2689 sort it - from lanshui_Yang

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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

06
    题目大意：给你一个数n ，然后有1 ~ n 的一个排列，让你找出这个排列的逆序数。
    解题思路：此题可以用树状数组来解，树状数组的三个用途：1.单点更新，区间求和 2、区间更新，单点求和
3、求逆序数。求逆序数想法较简单，请看代码：
#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<cmath>#include<algorithm>#include<queue>using namespace std ;const int MAXN = 1e5 + 7 ;int C[MAXN] ;int n ;int lowbit(int x){    return x & -x ;}int sum(int x){    int sumt = 0 ;    while (x > 0)    {        sumt += C[x] ;        x -= lowbit(x) ;    }    return sumt ;}void add(int x , int d){    while (x <= n)    {        C[x] += d ;        x += lowbit(x) ;    }}int main(){    while (scanf("%d" , &n) != EOF)    {        int i ;        int ans = 0 ;        memset(C , 0 ,sizeof(C)) ;        for(i = 1 ; i <= n ; i ++)        {            int a ;            scanf("%d" , &a) ;            add(a , 1) ;  // 此处是整个程序的精华部分，请好好理解            ans += i - sum(a) ;  // 统计逆序数        }        printf("%d\n" , ans) ;    }    return 0 ;}