UVA11014

  给定一个NxNxN的正方体，求出最多能选几个整数点。使得随意两点PQ不会使PQO共线。

 

  F(k)

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=5e5;
#define int long long
 int b[N+2], pm[N+2],tot=0;
 int n;
 int pow3(int x){
 	return x*x*x-1;
 }
 void init(){
   b[1] = 1;
    for (long long i = 2; i < N; i++) {
	if (b[i]) continue;
	pm[++tot] = i;
	for (long long j = i * i; j < N; j += i)
	    b[j] = 1;
    }
 }
int count(long long num) {
    int ans = 0;
    for (int i = 1; i <=tot && pm[i] <= num; i++) {
		if (!b[num]) {ans++; break;}
		if (num % pm[i] == 0) {
		    ans++;
		    num /= pm[i];
		    if (num % pm[i] == 0) return -1;
		}
    }
    return ans;
}
 long long cal(long long num) {
    int t = count(num);
    if (t == -1) return 0;
    if (t&1) return -pow3((n / 2 / num) * 2 + 1);
    else return pow3((n / 2 / num) * 2 + 1);
}

 signed main(){
 	init();
 	
 	int cas=0;
 	while(cin>>n&&n){
 		long long ans = pow3(n + 1);
    	for (long long i = 2; i <= n; i++)
			ans += cal(i);
 		printf("Crystal %d: %lld\n", ++cas, ans);	
 	}
 }