[hdu6428]Problem C. Calculate

题目大意：有$T(1leqslant Tleqslant 10)$组数据，每组数据给你$A,B,C(0<A,B,Cleqslant 10^7)$，求$sumlimits_{i=1}^Asumlimits_{j=1}^Bsumlimits_{k=1}^Cvarphi((i,j^2,k^3))mod 2^{30}$

题解：
$$
def dsum{displaystylesumlimits}
egin{align*}
&dsum_{i=1}^Adsum_{j=1}^Bdsum_{k=1}^Cvarphi((i,j^2,k^3))\
=&dsum_{d=1}^Avarphi(d)dsum_{i=1}^Adsum_{j=1}^Bdsum_{k=1}^C[(i,j^2,k^3)=d]\
end{align*}
$$

$$
def dsum{displaystylesumlimits}
令f(x)=dsum_{i=1}^Adsum_{j=1}^Bdsum_{k=1}^C[(i,j^2,k^3)=x]\
egin{align*}
F(x)&=dsum_{x|p}f(p)\
&=dsum_{i=1}^Adsum_{j=1}^Bdsum_{k=1}^C[x|(i,j^2,k^3)]\
&=dsum_{i=1}^A[x|i]dsum_{j=1}^B[x|j^2]dsum_{k=1}^C[x|j^3]\
&莫比乌斯反演得：\
f(x)&=dsum_{x|k}mu(dfrac k x)F(k)\
&=dsum_{i=1}^Amu(i)F(ix)\
ans&=dsum_{d=1}^Avarphi(d)dsum_{i=1}^Amu(i)F(id)\
&=dsum_{T=1}^AF(T)dsum_{d|T}varphi(d)mu(dfrac T d)\
&由狄利克雷卷积得：\
ans&=dsum_{T=1}^AF(T)(mu*varphi)(d)
end{align*}
$$

$$
狄利克雷卷积得(mu*varphi)(d)为积性函数\
def dsum{displaystylesumlimits}
令g(x)=dsum_{d|T}mu(d)varphi(dfrac T d)\
egin{align*}
g(1)&=1\
g(p)&=mu(1)varphi(p)+mu(p)varphi(1)\
&=1cdot(p-1)+(-1)cdot1\
g(p^k)&=mu(1)varphi(p^k)\
&+mu(p)varphi(p^{k-1})\
&qquadvdots\
&+mu(p^k)varphi(1)\
ecause&mu(p^k)当kgeqslant2时为0\
herefore g(p^k)&=mu(1)varphi(p^k)+mu(p)varphi(p^{k-1})\
&=p^k-k^{k-1}-(p^{k-1}-p^{k-2})\
&=p^k-2p^{k-1}+p^{k-2}\
herefore g(p^k)&=
egin{cases}
1(k=0)\
p-2(k=1)\
(p-1)^2(k=2)\
pcdot g(p^{k-1})(kgeqslant2)\
end{cases}
end{align*}\
可以用线性筛来做
$$

$$
def dsum{displaystylesumlimits}
def dprod{displaystyleprodlimits}
F(x)=dsum_{i=1}^A[x|i]dsum_{j=1}^B[x|j^2]dsum_{k=1}^C[x|j^3]\
易得dsum_{i=1}^A[x|i]=leftlfloordfrac A x ight floor\
考虑dsum_{j=1}^B[x|j^2]:\
对x分解质因数\
令x=dprod p_i^{c_i}\
令y_2(x)=dprod p_i^{leftlceildfrac{c_i}{2} ight ceil}\
x|j^2Rightarrow[y_2(x)|j]\
herefore dsum_{j=1}^B[x|j^2]=leftlfloordfrac{B}{y_2(x)} ight floor\
同理，令y_3(x)=dprod p_i^{leftlceildfrac{c_i}{3} ight ceil}\
herefore dsum_{k=1}^C[x|j^3]=leftlfloordfrac{B}{y_3(x)} ight floor\
herefore F(x)=leftlfloordfrac A x ight floorleftlfloordfrac{B}{y_2(x)} ight floorleftlfloordfrac{B}{y_3(x)} ight floor\
y_2(x),y_3(x)都可以线性筛
$$

卡点：无

C++ Code：

#include <cstdio>
#define maxn 10000010
#define mod 1073741823
int Tim, A, B, C;
int pl[maxn], ptot, g[maxn], f2[maxn], f3[maxn];
int cnt[maxn];
bool isp[maxn];
inline int sqr(int x) {return x * x;}
void sieve(int n) {
	g[1] = f2[1] = f3[1] = 1;
	for (int i = 2; i < n; i++) {
		if (!isp[i]) {
			pl[ptot++] = i;
			g[i] = i - 2;
			f2[i] = f3[i] = i;
			cnt[i] = 1;
		}
		for (int j = 0; j < ptot && pl[j] * i < n; j++) {
			int t = pl[j] * i;
			isp[t] = true;
			if (i % pl[j] == 0) {
				cnt[t] = cnt[i] + 1;
				int p = i / pl[j];
				if (p % pl[j]) g[t] = g[p] * sqr(pl[j] - 1);
				else g[t] = g[i] * pl[j];
				f2[t] = f2[i] * (cnt[t] & 1 ? pl[j] : 1);
				f3[t] = f3[i] * (cnt[t] % 3 == 1 ? pl[j] : 1);
				break;
			}
			cnt[t] = 1;
			g[t] = g[i] * g[pl[j]];
			f2[t] = f2[i] * f2[pl[j]];
			f3[t] = f3[i] * f3[pl[j]];
		}
	}
}
int main() {
	sieve(maxn);
	scanf("%d", &Tim);
	while (Tim --> 0) {
		scanf("%d%d%d", &A, &B, &C);
		int ans = 0;
		for (int i = 1; i <= A; i++) ans += g[i] * (A / i) * (B / f2[i]) * (C / f3[i]);
		printf("%d
", ans & mod);
	}
	return 0;
}