PAT-T1027 Larry and Inversions （樹狀數組）

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
int n;
int a[maxn];
int c[maxn];
int lowbit(int x) {
    return x&-x;
}
void add (int x,int val) {
    for (int i=x;i<=n;i+=lowbit(i)) c[i]+=val;
}
int getsum (int x) {
    int ans=0;
    for (int i=x;i;i-=lowbit(i)) ans+=c[i];
    return ans;
}
vector<int> ans;
int main () {
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",a+i);
    int wjm=0;
    for (int i=1;i<=n;i++) {
        wjm+=getsum(n)-getsum(a[i]);
        add(a[i],1);
    } 
    for (int i=1;i<=n;i++) {
        memset(c,0,sizeof(c));
        int sum=0;
        int sum_=0;//順序對數量 
        for (int j=i;j<=n;j++) {
            sum+=getsum(n)-getsum(a[j]);
            sum_+=getsum(a[j]);
            //翻轉一段區間，答案就是總的逆序對數量減去這個區間內的逆序對數量加上這個區間的順序對數量 
            ans.push_back(wjm-sum+sum_);
            add(a[j],1);
        }
        // 
    }
    for (int i=0;i<ans.size();i++) {
        if (i) printf(" ");
        printf("%d",ans[i]);
    }
}