题目
简化题意
(f(x) = sumlimits_{dmid x} d)，求(f(x) + f(x + 1) + dots + f(y)) 的值
思路。
整除分块。
一个数 (i) 在 (1 sim n) 中它的倍数一共有 (leftlfloor dfrac{n}{i} ight floor) 个，即 (i) 对答案的贡献为 (leftlfloor dfrac{n}{i} ight floor imes i)。
设 (g(n)) 表示 (sumlimits_{i = 1}^{n} f(i))，原式可以用下面的方式表示，
[egin{aligned}
g(y) - g(x - 1) &= sum_{i = 1}^{y}f(i) - sum_{j = 1}^{x - 1}f(j)\
&= sum_{i = 1} ^ {y} sumlimits_{cmid i} c - sum_{j = 1}^{j - 1} sumlimits_{dmid j} d
end{aligned}
]
考虑先枚举约数。
[egin{aligned}
&sum_{i = 1} ^ {y} sumlimits_{cmid i} c - sum_{j = 1}^{x - 1} sumlimits_{dmid j} d \
&= sum_{i = 1}^{y} leftlfloor dfrac{y}{i}
ight
floor imes i - sum_{j = 1}^{x - 1} leftlfloor dfrac{x - 1}{j}
ight
floor imes j
end{aligned}
]
可以用整除分块 (O(2sqrt n)) 求出答案。
Code
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
long long solve(int x) {
long long ans = 0;
for (int i = 1, j = 0; i <= x; i = j + 1) {
j = x / (x / i);
ans += 1ll * (x / i) * (1ll * (j - i + 1) * (1ll * j + i) / 2);
}
return ans;
}
int main() {
int x, y;
std::cin >> x >> y;
std::cout << solve(y) - solve(x - 1);
return 0;
}