Description:
共(T le 5 imes 10^4)组询问, 每组询问给定(n)和(m), 请你求出
[sum_{i = 1}^n sum_{j = 1}^m sigma_0 (ij)
]
Solution:
先给出一个结论:
[sigma_0(ij) = sum_{a | i} sum_{b | j} [gcd(a, b) = 1]
]
证明: 我们令(i = p_1^{a_1} p_2^{a_2} cdots), (j = p_1^{b_1} p_2^{b_2} cdots), (d | ij)且(d = p_1^{c_1} p_2^{c_2} cdots), 则(c_n le a_n + b_n).
考虑如何不重复地统计每一个(d): 令(c_n = A_n + B_n), 其中(A_n)和(B_n)分别为(i)和(j)对(c_n)的贡献, 则我们要求
[egin{cases}
B_n = 0 & A_n < a_n \
B_n ge 0 & A_n = a_n
end{cases}
]
这样一来, (c_n)的表示形式就变成唯一的了, 因而不会被重复统计. 我们再考虑如何统计这样的(A_n)和(B_n): 我们令(A_n' = a_n - A_n), 则约束条件变为
[egin{cases}
B_n = 0 & A_n'
e 0 \
B_n ge 0 & A_n' = 0
end{cases}
]
等价于(gcd(A_n', B_n) = 1).
因此得证.
好吧, 假如看不懂上面的这一些证明, 就这么想吧: (i)表示(a)中不取多少, (j)表示(b)中取多少, 只要保证(gce(a, b) = 1), 即不会重复统计.
因此我们考虑原题的式子
[egin{aligned}
sum_{i = 1}^n sum_{j = 1}^m sigma_0(ij) &= sum_{i = 1}^n sum_{j = 1}^m sum_{a | i} sum_{b | j} [gcd(a, b) = 1] \
&= sum_{i = 1}^n sum_{j = 1}^m sum_{a | i} sum_{b | j} sum_{d | gcd(a, b)} mu(d) \
&= sum_{i = 1}^n sum_{j = 1}^m sum_{d | gcd(i, j)} mu(d) sigma_0(frac i d) sigma_0(frac j d) \
&= sum_{d = 1}^n sum_{i = 1}^{lfloor frac n d
floor} sum_{j = 1}^{lfloor frac m d
floor} mu(d) sigma_0(i) sigma_1(j) \
&= sum_{d = 1}^n mu(d) sum_{i = 1}^{lfloor frac n d
floor} sigma_0(i) sum_{j = 1}^{lfloor frac m d
floor} sigma_0(j)
end{aligned}
]
分块处理后半部分即可.
时间复杂度: 预处理(O(n)), 单次询问(O(n^frac 1 2))
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
using namespace std;
namespace Zeonfai
{
inline int getInt()
{
int a = 0, sgn = 1; char c;
while (! isdigit(c = getchar())) if (c == '-') sgn *= -1;
while (isdigit(c)) a = a * 10 + c - '0', c = getchar();
return a * sgn;
}
}
const int N = (int)5e4, MOD = (int)1e9;
typedef int arr[N + 7];
typedef long long Larr[N + 7];
int tot;
arr isNotPrime, prm, mu, minDivisor, minDivisorDegree, sgm;
Larr a, b, c;
inline void initialize()
{
memset(isNotPrime, 0, sizeof(isNotPrime));
tot = 0;
sgm[1] = mu[1] = 1;
for (int i = 2; i <= N; ++ i)
{
if (! isNotPrime[i])
{
prm[tot ++] = i;
mu[i] = -1;
minDivisor[i] = i;
minDivisorDegree[i] = 1;
sgm[i] = 2;
}
for (int j = 0; j < tot && i * prm[j] <= N; ++ j)
{
int x = i * prm[j]; isNotPrime[x] = 1;
if (i % prm[j])
{
mu[x] = - mu[i];
minDivisor[x] = prm[j];
minDivisorDegree[x] = 1;
sgm[x] = sgm[i] * 2;
}
else
{
mu[x] = 0;
minDivisor[x] = minDivisor[i] * prm[j];
minDivisorDegree[x] = minDivisorDegree[i] + 1;
sgm[x] = sgm[i / minDivisor[i]] * (minDivisorDegree[x] + 1);
}
}
}
a[0] = b[0] = c[0] = 0;
for (int i = 1; i <= N; ++ i) a[i] = a[i - 1] + sgm[i], b[i] = a[i] * a[i], c[i] = c[i - 1] + mu[i];
}
int main()
{
using namespace Zeonfai;
initialize();
int T = getInt();
for (int cs = 0; cs < T; ++ cs)
{
int n = getInt(), m = getInt();
long long ans = 0;
int L = 1;
while (L <= min(n, m))
{
int R = min(n / (n / L), m / (m / L));
ans = ans + a[n / L] * a[m / L] * (c[R] - c[L - 1]);
L = R + 1;
}
printf("%lld
", ans);
}
}