树状数组模板3——求逆序对

树状数组的模板，通过其性质，寻找i<j但a[i]>a[j]的个数，此处添加了离散化来优化时间，减小常数

#include<cstdio>
#include<cctype>
#include<algorithm>
using namespace std;
 
int c[100010],n,tot,b[100010];
long long ans;
struct node
{
    int x,val;
    bool operator <(const node &k)const{
        return val<k.val;
    }
}a[100010];
 
int read()
{
    int x=0,f=1;
    char c=getchar();
    while (!isdigit(c))
        f=c=='-'?-1:1,c=getchar();
    while (isdigit(c))
        x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
 
int lowbit(int x)
{
    return x&-x;
}
 
int query(int x)
{
    int tot=0;
    for (int i=x;i;i-=lowbit(i))
        tot+=c[i];
    return tot;
}
 
void update(int x,int val)
{
    for (int i=x;i<=n;i+=lowbit(i))
        c[i]+=val;
}
 
int main()
{
    int i,j;
    n=read();
    for (i=1;i<=n;i++)
    {
        a[i].x=i;
        a[i].val=read();
    }
    sort(a+1,a+n+1);
    for (i=1;i<=n;i++)
    {
        if (i==1||a[i].val!=a[i-1].val)
            tot++;
        b[a[i].x]=tot;
    }
    for (i=1;i<=n;i++)
    {
        ans+=i-1-query(b[i]);
        update(b[i],1);
    }
    printf("%lld",ans);
    return 0;
}