B

考虑分类讨论：

1. $k=1k\; =\; 1$，高斯求和公式即可。
2. $k=2k\; =\; 2$，平方和公式，${\sum }_{i=1}^n={\displaystyle \frac{n(n+1)(2n+1)}{6}}\backslash sum\backslash limits\_\{i=1\}^\{n\}=\backslash dfrac\{n(n+1)(2n+1)\}\{6\}$。
3. $k\ge 3k\; \backslash ge\; 3$，由于 $r\le 10^{18}r\backslash le10^\{18\}$，直接枚举所有可行情况，类似整除分块处理即可。

整除分块：把所有值相同的考虑到一起求和，时间复杂度 $O(\sqrt[k]{10^{18}})O(\backslash sqrt[k]\{10^\{18\}\})$。

#include<cmath>
#include<cstdio>
#define int __int128
int init(){
	char c = getchar();
	int x = 0, f = 1;
	for (; c < '0' || c > '9'; c = getchar())
		if (c == '-') f = -1;
	for (; c >= '0' && c <= '9'; c = getchar())
		x = (x << 1) + (x << 3) + (c ^ 48);
	return x * f;
}
void print(int x){
	if (x < 0) x = -x, putchar('-');
	if (x > 9) print(x / 10);
	putchar(x % 10 + '0');
}
int sqr(int x){
	return x * (x + 1) * (2 * x + 1) / 6;
}
int sum(int l, int r){
	if (l > r) return 0;
	int L = sqr(l - 1);
	int R = sqr(r);
	return ((l + r) * (r - l + 1) / 2) + 2 * (R - L);
}
int quick_mod(int a, int b){
	int s = 1;
	while (b) {
		if (b & 1) s = s * a;
		a = a * a; b >>= 1;
		if (!b) break;
		if (a > (int) 1e18) return (int) 1e18;
		if (s > (int) 1e18) return (int) 1e18;
	}
	return s;
}
int mn(int x, int y){
	return x < y ? x : y;
}
int mx(int x, int y){
	return x > y ? x : y;
}
signed main(){
	int l = init(), r = init(), k = init();
	if (k == 1) {
		print((l + r) * (r - l + 1) / 2);
		return 0;
	}
	else if (k == 2) {
		int i = (int) sqrt((double) l);
		int j = (int) sqrt((double) r);
		int s = sum(i + 1, j - 1);
		print(s + (i * i + 2 * i - l + 1) * i + (r - j * j + 1) * j);
		return 0;
	}
	else {
		int ans = 0;
		for (int i = 1; i; ++i) {
			int L = quick_mod(i, k);
			int R = quick_mod(i + 1, k) - 1;
			if (R < l) continue;
			if (L > r) break;
			int num = mn(r, R) - mx(l, L) + 1;
			ans += i * num;
		}
		print(ans), putchar('\n');
		return 0;
	}
}