BZOJ1257 餘數之和 整除分塊

題目鏈接

https://www.lydsy.com/JudgeOnline/problem.php?id=1257

分析

$k \ \ mod \ \ i = k - i \times \lfloor {k \over i} \rfloor$，則要求 $n \times k - \sum_{i = 1}^n i \times \lfloor {k \over i} \rfloor$。

套用整除分塊，注意 $k$ 和 $n$ 的大小。

AC代碼

#include <cstdio>
#include <algorithm>

using namespace std;

inline int read() {
	int num = 0;
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	while (c >= '0' && c <= '9')
		num = num * 10 + c - '0', c = getchar();
	return num;
}

int main() {
	int n = read(), k = read();
	long long ans = 1ll * n * k;
	for (int l = 1, r; l <= n; l = r + 1) {
		r = k / l ? min(k / (k / l), n) : n;
		ans -= 1ll * (l + r) * (r - l + 1) / 2 * (k / l);
	}
	printf("%lld", ans);
	return 0;
}