B. Magic Forest
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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
Copy
6
output
Copy
1
input
Copy
10
output
Copy
2
Note
The only xorangle in the first sample is (3, 5, 6).
题意：问在1~n的范围内，有多少对构成三角形的边满足下列条件1 ≤ a ≤ b ≤ c ≤ n;a^b^c=0；
思路：由a^b^c=0我们可以枚举a,b；c=a^b;暴力跑一发就行了；
下面附上我的代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
bool check(int a,int b,int c){
if(a+b>c&&a+c>b&&b+c>a&&a-b<c&&b-a<c&&a-c<b&&c-a<b&&c-b<a&&b-c<a){
return true;
}else{
return false;
}
}
int main()
{
int n;
scanf("%d", &n);
int i, j;
int num=0;
for(i=1;i<=n;i++){
for(j=i;j<=n;j++){
int k = i ^ j;
if(check(i,j,k) && k >= j && k <=n){
num++;
}
}
}
printf("%d\n",num);
return 0;
}