【HDU - 3333】【Turing Tree】

题目：

After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new problems about intervals. So, today, this sick thing happens again... 

Now given a sequence of N numbers A1, A2, ..., AN and a number of Queries(i, j) (1≤i≤j≤N). For each Query(i, j), you are to caculate the sum of distinct values in the subsequence Ai, Ai+1, ..., Aj.

Input

The first line is an integer T (1 ≤ T ≤ 10), indecating the number of testcases below. 

For each case, the input format will be like this: 

* Line 1: N (1 ≤ N ≤ 30,000). 

* Line 2: N integers A1, A2, ..., AN (0 ≤ Ai ≤ 1,000,000,000). 

* Line 3: Q (1 ≤ Q ≤ 100,000), the number of Queries. 

* Next Q lines: each line contains 2 integers i, j representing a Query (1 ≤ i ≤ j ≤ N).

Output

For each Query, print the sum of distinct values of the specified subsequence in one line.

Sample Input

2
3
1 1 4
2
1 2
2 3
5
1 1 2 1 3
3
1 5
2 4
3 5      

Sample Output

1
5
6
3
6      

题目分析：就是给完咱们每个点的值，在进行询问 ，每个区间内不同数字的和，看似很简单的问题，但是却没有用线段树写过，所以难度有点大，最后参考的题解，明确了解题思路，先把每个区间按右端点排序，在从右向左遍历，如果这个数字第一次出现，就加上，如果再次出现了这个数字，那么就把之前已经出现的数字删除，将新的数字添上去。

ac代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define maxn 100005
using namespace std;
typedef long long ll;

struct node{
	int l,r;
	int s;
}q[maxn];
bool cmp(node a,node b)
{
	return a.r<b.r;
}
ll bi[maxn],num[maxn];
ll c[maxn],ans[maxn];
int last[maxn];
int n;

int lowbit(int x)
{
	return x&-x;
}
void add(int x,int v)
{
	while(x<=n)
	{
		c[x]+=v;
		x+=lowbit(x);	
	}	
} 
ll sum(int i)
{
	ll ret=0;
	while(i>0)
	{
		ret+=c[i];
		i-=lowbit(i);
	}
	return ret;
}

int main()
{
	int t,Q;
	cin>>t;
	while(t--)
	{
		memset(c,0,sizeof(c));
		memset(last,0,sizeof(last));
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			scanf("%I64d",&num[i]);
			bi[i]=num[i];
		}
		scanf("%d",&Q);
		for(int i=1;i<=Q;i++)
		{
			cin>>q[i].l>>q[i].r;
			q[i].s=i;
		}
		sort(q+1,q+1+Q,cmp);
		sort(bi+1,bi+1+n);
		int k=1;
		for(int i=1;i<=n;i++)
		{
			int pos=lower_bound(bi+1,bi+1+n,num[i])-bi;
			if(!last[pos])
			{
				add(i,num[i]);
				last[pos]=i;
			}
			else
			{
				add(last[pos],-num[i]);
				add(i,num[i]);
				last[pos]=i;
			}
			while(q[k].r==i&&k<=Q)
			{
				ans[q[k].s]=sum(i)-sum(q[k].l-1);
				k++;
			}
		}
		for(int i=1;i<=Q;i++)
		{
			printf("%I64d\n",ans[i]);
		}
	}
	return 0;
}