Codeforces 235E
原题
题目描述：设(d(n))表示(n)的因子个数， 给定(a, b, c)， 求：
[sum_{i=1}^{a} sum_{j=1}^{b} sum_{k=1}^{c} d(i cdot j cdot k) (mod 2^{30})
]
solution
rng_58 Orz，这方法太神了，rng_58证明了下面这条式子:
[sum_{i=1}^{a} sum_{j=1}^{b} sum_{k=1}^{c} d(i cdot j cdot k) =sum_{(i, j)=(i, k)=(j, k)=1} left lfloor frac{a}{i}
ight
floor left lfloor frac{b}{j}
ight
floor left lfloor frac{c}{k}
ight
floor
]
证明：
设
[f(a, b, c)=sum_{i=1}^{a} sum_{j=1}^{b} sum_{k=1}^{c} d(i cdot j cdot k)
]
[g(a, b, c)=sum_{(i, j)=(i, k)=(j, k)=1} left lfloor frac{a}{i}
ight
floor left lfloor frac{b}{j}
ight
floor left lfloor frac{c}{k}
ight
floor
]
由容斥原理可得（一式）
[d(i cdot j cdot k)=f(a, b, c)-f(a-1, b, c)-f(a, b-1, c)-f(a, b, c-1)+f(a-1, b-1, c)+f(a-1, b, c-1)+f(a, b-1, c-1)-f(a-1, b-1, c-1)
]
则若（二式）
[d(i cdot j cdot k)=g(a, b, c)-g(a-1, b, c)-g(a, b-1, c)-g(a, b, c-1)+g(a-1, b-1, c)+g(a-1, b, c-1)+g(a, b-1, c-1)-g(a-1, b-1, c-1)
]
则原命题得证。
二式(=)
[sum_{(i, j)=(i, k)=(j, k)=1} left lfloor frac{a}{i}
ight
floor left lfloor frac{b}{j}
ight
floor left lfloor frac{c}{k}
ight
floor -left lfloor frac{a-1}{i}
ight
floor left lfloor frac{b}{j}
ight
floor left lfloor frac{c}{k}
ight
floor - left lfloor frac{a}{i}
ight
floor left lfloor frac{b-1}{j}
ight
floor left lfloor frac{c}{k}
ight
floor - left lfloor frac{a}{i}
ight
floor left lfloor frac{b}{j}
ight
floor left lfloor frac{c-1}{k}
ight
floor + left lfloor frac{a-1}{i}
ight
floor left lfloor frac{b-1}{j}
ight
floor left lfloor frac{c}{k}
ight
floor +
left lfloor frac{a-1}{i}
ight
floor left lfloor frac{b}{j}
ight
floor left lfloor frac{c-1}{k}
ight
floor + left lfloor frac{a}{i}
ight
floor left lfloor frac{b-1}{j}
ight
floor left lfloor frac{c-1}{k}
ight
floor - left lfloor frac{a-1}{i}
ight
floor left lfloor frac{b-1}{j}
ight
floor left lfloor frac{c-1}{k}
ight
floor]
(=)
[sum_{(i, j)=(i, k)=(j, k)=1} (left lfloor frac{a}{i}
ight
floor -left lfloor frac{a-1}{i}
ight
floor) (left lfloor frac{b}{j}
ight
floor - left lfloor frac{b-1}{j}
ight
floor) (left lfloor frac{c}{k}
ight
floor - left lfloor frac{c-1}{k}
ight
floor)
]
即只有当((i, j)=(i, k)=(j, k)=1 , i|a, j|b, k|c)时，和中的式子才等于(1)，否则为(0).
设(p_i)为质因子，(q_i)为(p_{i}^{q_i} leq n)的最大值，则(n)的因数个数为
[prod_{i} (q_i +1)
]
根据上述定义设类似(q_i)的定义对于(a)为(x_i), (b)为(y_i)， (c)为(z_i)
对于(p_i)，该质数的个数为(x_i+y_i+z_i),
因为((i, j)=(i, k)=(j, k)=1 , i|a, j|b, k|c)， 对于(p_i), 答案为((0, 0, 0)+(1 ext ~ x_i, 0, 0)+(0, 1 ext ~ y_i, 0)+(0, 0, 1 ext ~ z_i)=x_i+y_i+z_i+1)
所以二式=一式，即(f(a, b, c)=g(a, b, c))
然后就可以用莫比乌斯的性质函数来解了。
[sum_{(i, j)=(i, k)=(j, k)=1} left lfloor frac{a}{i}
ight
floor left lfloor frac{b}{j}
ight
floor left lfloor frac{c}{k}
ight
floor
]
[=sum_{i} left lfloor frac{a}{i}
ight
floor sum_{d=(j, k)} epsilon(d) left lfloor frac{b}{j}
ight
floor left lfloor frac{c}{k}
ight
floor
]
[=sum_{i} left lfloor frac{a}{i}
ight
floor sum_{d} mu(d) left lfloor frac{b}{j'd}
ight
floor left lfloor frac{c}{k'd}
ight
floor
]
因为(i)与(d, j', k')都有关联，所以只好枚举
枚举(i),枚举(d)，然后分别枚举(j'), (k'),然后相乘，时间复杂度为：(O(n^2ln) (n))