HDU

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4980    Accepted Submission(s): 3431

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

冒泡排序要交换几次 模拟冒泡排一下记录即可 或者用树状数组求逆序数

①

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <vector>#define max_ 1010#define inf 0x3f3f3f3f#define ll long longusing namespace std;int n;int num[max_];int main(int argc, char const *argv[]){while(scanf("%d",&n)!=EOF){int i;for(i=1;i<=n;i++)scanf("%d",&num[i]);int cnt=0;for(i=1;i<n;i++){for(int j=1;j<n;j++){if(num[j]>num[j+1]){swap(num[j],num[j+1]);cnt++;}}}printf("%d\n",cnt);}return 0;}

②

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<set>#include<map>#include<iostream>#include<algorithm>using namespace std;#define N 100005#define maxn 10005#define ll long longint c[maxn],n;int lowbit(int a){    return a&(-a);}void update(int i,int x){    while(i<=n)    {        c[i]+=x;        i+=lowbit(i);    }}int getsum(int i){    int sum=0;    while(i>0)    {        sum+=c[i];        i-=lowbit(i);    }    return sum;}int main(){    int a[maxn];    while(cin>>n)    {        memset(c,0,sizeof(c));        int sum=0;        for(int i=0;i<n;i++)            {                cin>>a[i];                update(a[i],1);                sum+=i-getsum(a[i]-1);            }            cout<<sum<<endl;}return 0;}