n == 1的时候答案为区间长度， n == 2的时候每两个数字都可能成为答案，

我们只需要考虑 n == 3的情况， 我们可以枚举公差， 其分子分母都在sqrt(1e7)以内，

然后暴力枚举就好啦。

#include
     
      #define LL long long#define fi first#define se second#define mk make_pair#define PLL pair
      
       #define PLI pair
       
        #define PII pair
        
         #define SZ(x) ((int)x.size())#define ull unsigned long longusing namespace std;const int N = 1e5 + 7;const int inf = 0x3f3f3f3f;const LL INF = 0x3f3f3f3f3f3f3f3f;const int mod = 1e9 + 7;const double eps = 1e-6;const double PI = acos(-1);int n, l, r;struct Node {    int x, y;    bool operator < (const Node& rhs) const {        return x * rhs.y < rhs.x * y;    }    bool operator == (const Node& rhs) const {        return x * rhs.y == rhs.x * y;    }};vector
         
           vc;int Power(int a, int b) {    int ans = 1;    while(b) {        if(b & 1) ans = ans * a;        a = a * a; b >>= 1;    }    return ans;}int main() {    scanf("%d%d%d", &n, &l, &r);    if(n > 25) {        puts("0");    } else if(n == 1) {        printf("%d\n", r - l + 1);    } else if(n == 2) {        printf("%lld\n", 1ll * (r - l + 1) * (r - l));    } else {        for(int i = 1; i * i <= r; i++) {            for(int j = 1; j * j <= r; j++) {                if(i == j) continue;                int x = i, y = j;                int gcd = __gcd(x, y);                vc.push_back(Node{x / gcd, y / gcd});            }        }        sort(vc.begin(), vc.end());        vc.erase(unique(vc.begin(), vc.end()), vc.end());        LL ans = 0;        for(auto& t : vc) {            LL x = t.x, y = t.y;            bool flag = true;            for(int i = 3; i <= n; i++) {                y = y * t.y;                if(y > r) {                    flag = false;                    break;                }            }            if(!flag) continue;            for(int i = 3; i <= n; i++) {                x = x * t.x;                if(x > 10000000LL * y) {                    flag = false;                    break;                }            }            if(!flag) continue;            int Ly = ((l - 1) / y) + 1, Ry = r / y;            int Lx = ((l - 1) / x) + 1, Rx = r / x;            if(Ly > Ry) continue;            if(Lx > Rx) continue;            if(Lx > Ry) continue;            if(Rx < Ly) continue;            Lx = max(Lx, Ly);            Rx = min(Rx, Ry);            ans += Rx - Lx + 1;        }        printf("%lld\n", ans);    }    return 0;}/*3 1 10000000*/