【模板】樹狀數組求逆序對

具體學習參考https://blog.csdn.net/ssimple_y/article/details/53744096

#include<bits/stdc++.h>  
using namespace std;  
const int N=2e5+5;  
ll p,a[N],sum[N],s[N];  
int n;  
int lowbit(int i)  
{  
    return i&(-i);  
}  
void update(int i)  
{  
    while(i<=n)  
    {  
        sum[i]+=1;  
        i+=lowbit(i);  
    }  
}  
ll get_sum(ll i)  
{  
    ll ans=0;  
    while(i)  
    {  
        ans+=sum[i];  
        i-=lowbit(i);  
    }  
    return ans;  
}  
int main()  
{  
    while(~scanf("%d",&n))  
    {  
        memset(sum,0,sizeof(sum));  
        memset(s,0,sizeof(s));  
        ll ans=0,Ans=0;  
        for(int i=1; i<=n; i++)  
        {  
            scanf("%lld",&a[i]);  
            s[i]=a[i];  
        }  
        //離散化  
        sort(s,s+n);  
        int k=unique(s,s+n)-s;  
        for(int i=1; i<=n; i++) a[i]=(lower_bound(s,s+k,a[i])-s)+1;//保證下標從1開始  
  
        for(int i=1; i<=n; i++)  
        {  
            ll tmp=get_sum(a[i]);  
            update(a[i]);  
            ans+=tmp;//i前面小於等於a[i]的數個數  
            Ans+=i-1-tmp;//i前面小於等於a[i]的數個數，逆序數  
//        printf("i==%d a[i]==%lld tmp==%lld  ans==%lld\n",i,a[i],tmp,i-tmp);  
        }  
        printf("%lld\n",ans);//小於等於當前元素的個數總和  
        printf("%lld\n",Ans);//求逆序數  
    }  
    return 0;  
}