# [bzoj2301][HAOI2011]Problem b

\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\\ 莫比乌&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\\

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;
}


posted @ 2018-08-23 15:17  Memory_of_winter  阅读(72)  评论(0编辑  收藏