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.
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.
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 ; }