题干：
Imp is in a magic forest, where xorangles grow (wut?)
A xorangle of order n is such a non-degenerate triangle, that lengths of its sides are integers not exceeding n, and the xor-sum of the lengths is equal to zero. Imp has to count the number of distinct xorangles of order n to get out of the forest.
Formally, for a given integer n you have to find the number of such triples (a, b, c), that:
- 1 ≤ a ≤ b ≤ c ≤ n;
, where
denotes the bitwise xor of integers x and y.
- (a, b, c) form a non-degenerate (with strictly positive area) triangle.
Input
The only line contains a single integer n (1 ≤ n ≤ 2500).
Output
Print the number of xorangles of order n.
Examples
Input
6
Output
1
Input
10
Output
2
Note
The only xorangle in the first sample is (3, 5, 6）.
题目大意：
问你有多少个非严格递增的三个数a,b,c,满足a^b^c=0.
解题报告：
这题因为异或的性质，a^a=0，0^c=c。所以在三元组中，枚举前两小的值，那么第三个值其实就已知了。也就是这三个值一定是a , b , a^b。然后把题目那些条件带进去做判断就可以了
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int n;
int main()
{cin>>n;ll ans = 0;for(int i = 1; i<=n; i++) {for(int j = i; j<=n; j++) {if((i^j)<=n &&(i^j)>=j&&(i+j) > (i^j)) {//printf("%d %d\n",i,j);ans++;}}}printf("%lld\n",ans);return 0 ;}
