BZOJ 3309 DZY Loves Math

来源：互联网发布：淘宝网百雀羚旗舰店编辑：程序博客网时间：2024/05/17 18:25

题目链接

题意

求

\sum i = 1 a \sum j = 1 b f (g c d (i, j))

其中

f (x) = {α 1, x = p 1 α 1 * p 2 α 2 + . . . + p n α n, α 1 > α 2, . . ., α n 0, x = 1

推导

法一：艾弗森约定

记 gcd(i,j)=k, 则有 gcd(ik,jk)=1,
故原式可化为

\sum k = 1 m i n (a, b) \sum i = 1 ⌊ a k ⌋ \sum j = 1 ⌊ b k ⌋ f (k) [g c d (i, j) = 1]

用莫比乌斯函数将艾弗森约定展开得

\sum k = 1 m i n (a, b) f (k) \sum i = 1 ⌊ a k ⌋ \sum j = 1 ⌊ b k ⌋ \sum d | i, d | j μ (d)

即

\sum k = 1 m i n (a, b) f (k) \sum d = 1 m i n (⌊ a k ⌋, ⌊ b k ⌋) μ (d) \sum i = 1 ⌊ a k d ⌋ \sum j = 1 ⌊ b k d ⌋

即

\sum k = 1 m i n (a, b) f (k) \sum d = 1 m i n (⌊ a k ⌋, ⌊ b k ⌋) μ (d) ⌊ a k d ⌋ ⌊ b k d ⌋

令

T=kd, 得

\sum T = 1 m i n (a, b) ⌊ a T ⌋ ⌊ b T ⌋ \sum d | T f (T d) μ (d)

显然，接下来的路子就是求前缀和，然后分块计算，然而 关键问题 就是怎么求这个

\sum d | T f (T d) μ (d)

呢?

参考PoPoQQQ可知，对于

g (T) = \sum d | T f (T d) μ (d), T = p 1 α 1 * p 2 α 2 * . . . * p n α n

有

g (T) = {0, \exists α i \neq α j, (- 1) (n + 1), o t h e r w i s e

具体证明见上附链接，要点为从

k 个元素中选取奇数个元素的种数

= 选取偶数个元素的种数，因为

(1−1)k=0→C0k+C2k+...=C1k+C3k+...

法二：莫比乌斯反演（待补）

Code

#include <bits/stdc++.h>#define maxn 10000000#define maxm maxn + 10using namespace std;typedef long long LL;bool check[maxm];int prime[maxm], cnt[maxm], val[maxm], g[maxm], sum[maxm];void init() {    int tot = 0; g[1] = 0;    for (int i = 2; i <= maxn; ++i) {        if (!check[i]) {            prime[tot++] = i;            cnt[i] = 1, val[i] = i, g[i] = 1;        }        for (int j = 0; j < tot; ++j) {            if (i * prime[j] > maxn) break;            check[i * prime[j]] = true;            if (i % prime[j] == 0) {                cnt[i * prime[j]] = cnt[i] + 1, val[i * prime[j]] = val[i] * prime[j];                int temp = (i * prime[j]) / val[i * prime[j]];                if (temp == 1) g[i * prime[j]] = 1;                else g[i * prime[j]] = (cnt[temp] == cnt[i * prime[j]] ? -g[temp] : 0);                break;            }            cnt[i * prime[j]] = 1, val[i * prime[j]] = prime[j];            int temp = (i * prime[j]) / val[i * prime[j]];            if (temp == 1) g[i * prime[j]] = 1;            else g[i * prime[j]] = (cnt[temp] == cnt[i * prime[j]] ? -g[temp] : 0);        }    }    for (int i = 1; i <= maxn; ++i) sum[i] = sum[i - 1] + g[i];}void work() {    LL a, b;    scanf("%lld%lld", &a, &b);    int le, ri, lim = min(a, b);    LL ans = 0;    for (int i = 1; i <= lim; i = ri + 1) {        le = i, ri = min(a / (a / i), b / (b / i));        ans += (a / i) * (b / i) * (sum[ri] - sum[le - 1]);    }    printf("%lld\n", ans);}int main() {    init();    int T;    scanf("%d", &T);    while (T--) work();    return 0;}

阅读全文

0 0