[bzoj2301][HAOI2011]Problem b

题目大意：给你$a,b,c,d,k(1leqslant a,b,c,d,kleqslant 5 imes10^4)$，求$displaystylesumlimits_{x=a}^bdisplaystylesumlimits_{y=c}^d[(x,y)==k]$

题解：下文中令$nleqslant m$先考虑求$sumlimits_{i=1}^nsumlimits_{j=1}^m[(i,j)==k]$
$$
egin{align*}
defdsum{displaystylesumlimits}
令f(p)&=dsum_{i=1}^ndsum_{j=1}^m[(i,j)==p]\
令F(p)&=dsum_{p|k}f(k)\
    &=dsum_{p|k}dsum_{i=1}^ndsum_{j=1}^m[(i,j)==k]\
    &=dsum_{i=1}^ndsum_{j=1}^m[p|(i,j)]\
    &=leftlfloordfrac{n}{p} ight floorcdot leftlfloordfrac{m}{p} ight floor\
莫比乌&斯反演得：\
herefore f(p)&=dsum_{p|k}muBig(dfrac{k}{p}Big)F(k)\
            &=dsum_{i=1}^nmu(i)leftlfloordfrac{n}{ip} ight floorcdotleftlfloordfrac{m}{ip} ight floor\
end{align*}\
令g(p)=dsum_{i=1}^pmu(i)\
然后容斥一下就好了\
$$

卡点：无


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
", solve(b, d, k) - solve(a - 1, d, k) - solve(b, c - 1, k) + solve(a - 1, c - 1, k));
	}
	return 0;
}