題目大意：給你$a,b,c,d,k(1\leqslant a,b,c,d,k\leqslant 5\times10^4)$，求$\displaystyle\sum\limits_{x=a}^b\displaystyle\sum\limits_{y=c}^d[(x,y)==k]$

題解：下文中令$n\leqslant m$先考慮求$\sum\limits_{i=1}^n\sum\limits_{j=1}^m[(i,j)==k]$
$$
\begin{align*}
\def\dsum{\displaystyle\sum\limits}
令f(p)&=\dsum_{i=1}^n\dsum_{j=1}^m[(i,j)==p]\\
令F(p)&=\dsum_{p|k}f(k)\\
?? ?&=\dsum_{p|k}\dsum_{i=1}^n\dsum_{j=1}^m[(i,j)==k]\\
?? ?&=\dsum_{i=1}^n\dsum_{j=1}^m[p|(i,j)]\\
?? ?&=\left\lfloor\dfrac{n}{p}\right\rfloor\cdot \left\lfloor\dfrac{m}{p}\right\rfloor\\
莫比烏&斯反演得：\\
\therefore f(p)&=\dsum_{p|k}\mu\Big(\dfrac{k}{p}\Big)F(k)\\
?? ??? ??? ?&=\dsum_{i=1}^n\mu(i)\left\lfloor\dfrac{n}{ip}\right\rfloor\cdot\left\lfloor\dfrac{m}{ip}\right\rfloor\\
\end{align*}\\
令g(p)=\dsum_{i=1}^p\mu(i)\\
然后容斥一下就好了\\
$$

卡點：無


C++ Code：

#include <cstdio> #define maxn 50010 using namespace std; int miu[maxn], plist[maxn], ptot; bool isp[maxn]; void sieve(int n) {miu[1] = 1;for (int i = 2; i <= n; i++) {if (!isp[i]) {miu[i] = -1;plist[ptot++] = i;}for (int j = 0; j < ptot, i * plist[j] <= n; j++) {int tmp = i * plist[j];isp[tmp] = true;if (i % plist[j] == 0) {miu[tmp] = 0;break;}miu[tmp] = -miu[i];}}for (int i = 2; i <= n; i++) miu[i] += miu[i - 1]; } inline int min(int a, int b) {return a < b ? a : b;} int solve(int n, int m, int k) {n /= k, m /= k;int tmp = min(n, m);int ans = 0, l, r;for (l = 1; l <= tmp; l = r + 1) {r = min(n / (n / l), m / (m / l));ans += (miu[r] - miu[l - 1]) * (n / l) * (m / l);}return ans; } int Tim, a, b, c, d, k; int main() {sieve(50000);scanf("%d", &Tim);while (Tim --> 0) {scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);printf("%d\n", solve(b, d, k) - solve(a - 1, d, k) - solve(b, c - 1, k) + solve(a - 1, c - 1, k));}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/Memory-of-winter/p/9523943.html

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