Tree-like array - (discretization) find the number of reverse ordinal numbers of relatively large numbers

//树状数组大范围求逆序数的离散化模板    --cyl 
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100005;
int n;
int discrete[N],s[N];    //离散数组 
struct node
{
    long long  x;
    int num;    //下标 
}T[N]; 
bool cmp(node s1,node s2){
    return s1.x <s2.x;
}
int lowbit(int x){
    return x&(-x);
}
int sum(int x)  //求出现在x这个位置前边的和 这个求出是顺序的 
{
    int ans=0;
    while(x){
        ans+=s[x];
        x-=lowbit(x); 
    }
    return ans;
}
void insert(int x) //离散化后的数组更新 
{
    while(x<=n){
        s[x]++;
        x+=lowbit(x); 
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(s,0,sizeof(s));
        for(int i=1;i<=n;i++){
            scanf("%lld",&T[i].x);
            T[i].num=i;
        }
        sort(T+1,T+1+n,cmp);
        for(int i=1;i<=n;i++)
            discrete[T[i].num]=i;       //离散化后的坐标 
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            insert(discrete[i]);        //第i次输入后的位置 
            ans+=(i-sum(discrete[i]));  //所有的逆序数之和 
        } 
        printf("%d\n",ans); 
    } 
}