Sort it
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4110 Accepted Submission(s): 2920
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.
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
6
Author
WhereIsHeroFrom
Source
Recommend
思路：求出每个数ai前面有多少个数比ai大，或者每个数ai后面有多少个数比ai小。第一种方法只需要交换树状数组更新和求和的函数。第二种方法就是求逆向对的数量和只需要逆向输入a，直接标准的树状数组。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=;
int c[MAXN];
inline int Lowbit(int x)
{
return x&(-x);
}
void update(int i,int val)
{
for(i; i>; i-=Lowbit(i))
c[i]+=val;
}
int sum(int i)
{
int temp=;
for(i; i<=MAXN; i+=Lowbit(i))
temp+=c[i];
return temp;
}
int main()
{
int i,n;
int a;
while(~scanf("%d",&n))
{
int ans=;
memset(c,,sizeof(c));
for(i=; i<=n; i++)
{
scanf("%d",&a);
ans+=sum(a);
update(a,);
}
cout<<ans<<endl;
}
return ;
}
求左边大于等于 a[i]的数的个 数