Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1940    Accepted Submission(s): 1390

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

3 1 2 3 4 4 3 2 1

 

 

Sample Output

0 6

 

 

Author

WhereIsHeroFrom

 

 

Source

 

 

Recommend

yifenfei

 

题意是给你一个序列，问你最少交换多少次可以使整个序列单调上升。

数据只有1000，直接n^2可做，但是有一种更为精妙的方法——树状数组 0Ms

我们先初始化整个序列为0，然后每输入一个数x，就在x位置将0换成1，此时求该位置的sum，就是比x小的数字数目，显然i - sum就是是这个数字要交换的次数。

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;int a[1010], c[1010] , n;int lowbit(int x){    return x & (-x);}int sum(int x){    int sum = 0;    while(x > 0)    {        sum += c[x];        x -= lowbit(x);    }    return sum;}void update(int x , int num){    while(x <= n)    {        c[x] += num;        x += lowbit(x);    }}int main(){    while(~scanf("%d" , &n))    {        memset(a , 0 , sizeof(a));        memset(c , 0 , sizeof(c));        int ans = 0;        int x;        for(int i = 1 ; i <= n ; i++)        {            scanf("%d" , &x);            update(x , 1);            ans += i - sum(x);        }        printf("%d\n" , ans);    }    return 0;}