P2260 [清华集训2012]模积和，P2834 能力测验（二维除法分块）
P2260 [清华集训2012]模积和
推导过程
我们假定 n < = m n <= m n<=m
∑ i = 1 n ∑ j = 1 m ( n m o d i ) ( m m o d j ) , i ≠ j \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} (n\mod i)(m \mod j), i \not= j i=1∑nj=1∑m(nmodi)(mmodj),i=j
= ∑ i = 1 n ∑ j = 1 m ( n m o d i ) ( m m o d j ) − ∑ k = 1 n ( n m o d k ) ( m m o d k ) = \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} (n\mod i)(m \mod j) - \sum_{k = 1} ^{n} (n \mod k) (m \mod k) =i=1∑nj=1∑m(nmodi)(mmodj)−k=1∑n(nmodk)(mmodk)
= ∑ i = 1 n ( n − ⌊ n i ⌋ i ) ∑ j = 1 m ( m − ⌊ m j ⌋ j ) − ∑ k = 1 n ( n − ⌊ n k ⌋ k ) ( m − ⌊ m k ⌋ k ) = \sum_{i = 1} ^{n} (n - \lfloor \frac{n}{i}\rfloor i) \sum_{j = 1} ^{m} (m - \lfloor \frac{m}{j}\rfloor j) - \sum_{k = 1} ^{n} (n - \lfloor \frac{n}{k}\rfloor k) (m - \lfloor \frac{m}{k}\rfloor k) =i=1∑n(n−⌊in⌋i)j=1∑m(m−⌊jm⌋j)−k=1∑n(n−⌊kn⌋k)(m−⌊km⌋k)
= ( n 2 − ∑ i = 1 n ⌊ n i ⌋ i ) ( m 2 − ∑ j = 1 m ⌊ m j ⌋ j ) − ∑ k = 1 n ( n m − n k ⌊ m k ⌋ − m k ⌊ n k ⌋ ) + k 2 ⌊ n k m k ⌋ = (n ^ 2 - \sum_{i = 1} ^{n} \lfloor \frac{n}{i}\rfloor i)(m ^ 2 - \sum_{j = 1} ^{m} \lfloor \frac{m}{j}\rfloor j) - \sum_{k = 1} ^{n}(nm - nk \lfloor{\frac{m}{k}}\rfloor - mk \lfloor{\frac{n}{k}}\rfloor) + k ^ 2 \lfloor{\frac{n}{k}}\frac{m}{k}\rfloor =(n2−i=1∑n⌊in⌋i)(m2−j=1∑m⌊jm⌋j)−k=1∑n(nm−nk⌊km⌋−mk⌊kn⌋)+k2⌊knkm⌋
之后我们可以利用整除分块加平方和公式与逆元结合求得最后答案
1 + 2 2 + 3 2 + … … + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 1 + 2 ^ 2 + 3 ^ 2 + …… + n ^ 2 = \frac{n(n + 1)(2n + 1)}{6} 1+22+32+……+n2=6n(n+1)(2n+1)
多余的化简，只要上面的就行了，一开始我化简成了下面的式子，以为更简单，结果求崩了，找不出 b u g bug bug，，，
= ( n 2 − ∑ i = 1 n ⌊ n i ⌋ i ) ( m 2 − ∑ j = 1 m ⌊ m j ⌋ j ) − n 2 m + ∑ k = 1 n ( n k ⌊ m k ⌋ + m k ⌊ n k ⌋ ) − k 2 ⌊ n k ⌋ ⌊ m k ⌋ = (n ^ 2 - \sum_{i = 1} ^{n} \lfloor \frac{n}{i}\rfloor i)(m ^ 2 - \sum_{j = 1} ^{m} \lfloor \frac{m}{j}\rfloor j) - n ^ 2m + \sum_{k = 1} ^{n}(nk \lfloor{\frac{m}{k}}\rfloor + mk \lfloor{\frac{n}{k}}\rfloor) - k ^ 2 \lfloor{\frac{n}{k}}\rfloor\lfloor\frac{m}{k}\rfloor =(n2−i=1∑n⌊in⌋i)(m2−j=1∑m⌊jm⌋j)−n2m+k=1∑n(nk⌊km⌋+mk⌊kn⌋)−k2⌊kn⌋⌊km⌋
= ( n 2 − ∑ i = 1 n ⌊ n i ⌋ i ) ( m 2 − ∑ j = 1 m ⌊ m j ⌋ j ) − n 2 m + n 2 ∑ k = 1 n k ⌊ m k ⌋ + n m ∑ k = 1 n k ⌊ n k ⌋ ) − ∑ k = 1 n k 2 ⌊ n k ⌋ ⌊ m k ⌋ = (n ^ 2 - \sum_{i = 1} ^{n} \lfloor \frac{n}{i}\rfloor i)(m ^ 2 - \sum_{j = 1} ^{m} \lfloor \frac{m}{j}\rfloor j) - n ^ 2m + n ^ 2\sum_{k = 1} ^{n}k \lfloor{\frac{m}{k}}\rfloor + nm\sum_{k = 1} ^{n}k \lfloor{\frac{n}{k}}\rfloor) - \sum_{k = 1} ^{n}k ^ 2 \lfloor{\frac{n}{k}}\rfloor\lfloor\frac{m}{k}\rfloor =(n2−i=1∑n⌊in⌋i)(m2−j=1∑m⌊jm⌋j)−n2m+n2k=1∑nk⌊km⌋+nmk=1∑nk⌊kn⌋)−k=1∑nk2⌊kn⌋⌊km⌋
代码
注意随手取模，容易溢出 w a wa wa。
/*
Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include P2834 能力测验
照搬上面的代码，改一下模数和逆元即可。
/*
Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include