HDU 2689 sort it（树状数组 逆序数）

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

 

为什么可以用逆序数呢，因为它要相邻两个交换，而逆序数，比如 4 3 2 1，比3大的1个，正好4，3交换，比2大的2个，4和3，2先和3换，再和4换。就算4和3先不进行交换，2与3交换，变成4 2 3 1，在3前面比3大的还是4，还是会通过2，4交换，然后再3，4交换，所以这题可以通过求逆序数解

AC代码：

#include<cstdio>
#include<cstring>
const int N = 100000+5;
int n;
int c[N],a[N];

int lowbit(int x) {
    return x&-x;
}

void add(int x, int a) {
    while(x <= n) {
        c[x] += a;
        x += lowbit(x);
    }
}

int sum(int x) {
    int ans = 0;
    while(x) {
        ans += c[x];
        x -= lowbit(x);
    }
    return ans;
}

int main(){
    while(scanf("%d",&n)!=EOF) {
        memset(c,0,sizeof(c));    
        int ans=0;
        for(int i = 1; i <= n; i++) {
            scanf("%d",&a[i]);
            add(a[i], 1);
            ans += i-sum(a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}