Sort it hdu 2689

来源：互联网发布：e4a操作数据库编辑：程序博客网时间：2024/05/21 11:19

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
简单的一个水题，第一次看到的时候 直接暴力俩for循环计数下就ac了。
。 但是 后来用了树状数组来写，发现真的很神奇。理解下题意。就是看
输入的时候的数前面有几个逆序数，存入数组，然后  求1到n的逆序数个数。
注意输入的时候是先更新 ，防止最后一个未算入。
暴力写法：
#include<iostream>#include<cstring>#include<cstdlib>#include<string>using namespace std;int main(){int n,i,j;int s[1010];while(cin>>n){for(i=0;i<n;i++){cin>>s[i];}int tot=0;for(i=0;i<n;i++){for(j=i+1;j<n;j++){if(s[i]>s[j]){int t=s[i];s[i]=s[j];s[j]=t;tot++;}}}cout<<tot<<endl;}return 0;}

树状数组写法

#include<iostream>#include<cstring>using namespace std;int s[1010],n;int lowbit(int x){return x&(-x);}int sum(int x){int suM=0;while(x>0){suM+=s[x];x-=lowbit(x);}return suM;}void updata(int x){while(x<=n){++s[x];x+=lowbit(x);}}int main(){int i,m;while(cin>>n){memset(s,0,sizeof(s));int ans=0;for(i=1;i<=n;i++){cin>>m;updata(m);ans+=sum(n)-sum(m);}cout<<ans<<endl;}return 0;}

0 0