线段树求逆序对

题目链接 hdu-5775

P is a permutation of the integers from 1 to N(index starting from 1). 
Here is the code of Bubble Sort in C++. 

for(int i=1;i<=N;++i)
    for(int j=N,t;j>i;—j)
        if(P[j-1] > P[j])
            t=P[j],P[j]=P[j-1],P[j-1]=t;

After the sort, the array is in increasing order. ?? wants to know the absolute values of difference of rightmost place and leftmost place for every number it reached.

InputThe first line of the input gives the number of test cases T; T test cases follow. 
Each consists of one line with one integer N, followed by another line with a permutation of the integers from 1 to N, inclusive. 

limits 
T <= 20 
1 <= N <= 100000 
N is larger than 10000 in only one case. 
OutputFor each test case output “Case #x: y1 y2 … yN” (without quotes), where x is the test case number (starting from 1), and yi is the difference of rightmost place and leftmost place of number i.Sample Input

2
3
3 1 2
3
1 2 3

Sample Output

Case #1: 1 1 2
Case #2: 0 0 0

Hint

In first case, (3, 1, 2) -> (3, 1, 2) -> (1, 3, 2) -> (1, 2, 3)
the leftmost place and rightmost place of 1 is 1 and 2, 2 is 2 and 3, 3 is 1 and 3
In second case, the array has already in increasing order. So the answer of every number is 0.

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<cmath>
#include<string>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
ll c[100001],n,l[100001],r[100001],d[100001];
struct node{
    ll id,val,index;
    bool operator<(const node &th)const
    {
        return val<th.val; 
    }
}num[100001];
ll lowbit(ll x){
    return x&(-x);
}
void update(ll pos){
    while(pos<=n){
        c[pos]++;
        pos+=lowbit(pos);
    }
}
ll getsum(ll pos){
    ll sum=0;
    while(pos>0){
        sum+=c[pos];
        pos-=lowbit(pos);
    }
    return sum;
}

int main(){
    ll t,cnt=1;
    scanf("%lld",&t);
    while(t--){
        scanf("%lld",&n);
        memset(c,0,sizeof(c));
        for(ll i=1;i<=n;i++){
            scanf("%lld",&num[i].val);
            //cin>>num[i].val;
//            d[i]=num[i]
            l[num[i].val]=min(i,num[i].val);//最左边的位置
            num[i].id=n-i+1;//方便查看num[i].val后面比他小的个数
            num[i].index=i;
        }
        sort(num+1,num+1+n);
        for(ll i=1;i<=n;i++){
            ll mp=getsum(num[i].id);
            d[num[i].val]=num[i].index+mp-l[num[i].val];
            update(num[i].id);
        }
        printf("Case #%lld:",cnt++);
        for(ll i=1;i<=n;i++){
            printf(" %lld",d[i]);
        }
        printf("
");
    }
    return 0;
}