题意翻译
题目大意
给定一个正整数nn ，求满足如下条件的三元组(a,b,c)(a,b,c) 的个数：
- 1 le a le b le c le n1≤a≤b≤c≤n
- a space xor space b space xor space c=0a xor b xor c=0
- 存在一个边长分别为a,b,ca,b,c 的三角形。
输入格式
一行一个正整数n(1 le n le 2500)n(1≤n≤2500)
输出格式
输出满足题意的三元组个数。
感谢U3144 浮尘ii 提供的翻译
题目描述
Imp is in a magic forest, where xorangles grow (wut?)
A xorangle of order nn is such a non-degenerate triangle, that lengths of its sides are integers not exceeding nn , and the xor-sum of the lengths is equal to zero. Imp has to count the number of distinct xorangles of order nnto get out of the forest.
Formally, for a given integer nn you have to find the number of such triples (a,b,c)(a,b,c) , that:
- 1<=a<=b<=c<=n1<=a<=b<=c<=n ;
, where 
denotes the bitwise xor of integers xx and yy .- (a,b,c)(a,b,c) form a non-degenerate (with strictly positive area) triangle.
输入输出格式
输入格式：
The only line contains a single integer nn (1<=n<=2500)(1<=n<=2500) .
输出格式：
Print the number of xorangles of order nn .
输入输出样例
说明
The only xorangle in the first sample is (3,5,6)(3,5,6) .
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int n,ans; int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++){ int k=i^j; if(k>n) continue; if(i+j>k&&abs(i-j)<k&&i+k>j&&abs(i-k)<j&&j+k>i&&abs(j-k)<i) ans++; } cout<<ans/3; }