題目大意：
找出所有長度爲n，每一項都在l到r之間的等比數列的個數（q!=1)
解題思路：
其實除了幾種特殊情況外，基本都是枚舉分子分母互質的分數，考慮到只有兩項的時候數量較大，所以單獨判斷
代碼：
#include "iostream"
#include "cstdio"
#include "math.h"
#include "algorithm"
#include "string"
#include "string.h"
#include "vector"
#include "map"
#include "queue"
using namespace std;
long long n, l, r;
long long pow(long long a, long long x) {
long long ans = 1;
while (x) {
if (x & 1)
ans *= a;
a *= a;
x /= 2;
}
return ans;
}
long long gcd(long long a, long long b)
{
return (b>0) ? gcd(b, a%b) : a;
}
int main() {
while (scanf("%lld %lld %lld", &n, &l, &r) != EOF) {
if (n == 1) {
printf("%lld\n", r - l + 1);
}
else if (n == 2) {
printf("%lld\n", (r - l + 1)*(r - l));
}
else if (n > 25) {
puts("0");
}
else {
long long ans = 0;
long long maxnum = pow(2, log(100000000*0.1)/log(2.0) / (n - 1));
maxnum = min(maxnum, 3200LL);
for (long long i = 1;i <= maxnum;i++) {
for (long long j = i + 1;j <= maxnum;j++) {
if (gcd(i, j) == 1) {
long long a = pow(j, n - 1);
long long b = pow(i, n - 1);
if (l*a / b > r)
continue;
ans += (r*b / a) / b - (l - 1) / b;
}
}
}
printf("%lld\n", ans * 2);
}
}
return 0;
}