【51Nod 1363】最小公倍数之和

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1363

[egin{aligned} &sum_{i=1}^nfrac{in}{(i,j)}\ =&nsum_{d|n}sum_{i=1}^{frac nd}left[left(i,frac nd ight)=1 ight]i\ =&nleft(1+sum_{d|n,d eq n}sum_{i=1}^{leftlfloorfrac {n}{2d} ight floor}left[left(i,frac nd ight)=1 ight] ight)\ =&n+frac n2sum_{d|n,d eq 1}dvarphi(d) end{aligned} ]

重点是统计(sumlimits_{d|n,d eq 1}dvarphi(d))

[egin{aligned} &sum_{d|n,d eq 1}dvarphi(d)\ =&prodsum_{j=0}^{c_i}p_i^j imesvarphileft(p_i^j ight)-1\ =&prodleft(1+sum_{j=1}^{c_i}p_i^{2j-1}left(p_i-1 ight) ight)-1\ =&prodleft(1+frac{p_i^{2c_i+1}-p_i}{p_i+1} ight)-1 end{aligned} ]

质因子分解统计就可以了，时间复杂度(Oleft(Tsqrt n ight))。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;

const int N = 100000;
const int p = 1000000007;

bool notp[N + 3];
int T, n, prime[N + 3], num = 0, ni[N + 3];

void Euler_shai() {
	for (int i = 2; i <= N; ++i) {
		if (!notp[i]) prime[++num] = i;
		for (int j = 1; j <= num && prime[j] * i <= N; ++j) {
			notp[prime[j] * i] = true;
			if (i % prime[j] == 0) break;
		}
	}
	ni[1] = 1;
	for (int i = 2; i <= N; ++i)
		ni[i] = 1ll * (p - p / i) * ni[p % i] % p;
}

int ipow(int a, int b) {
	int r = 1, w = a;
	while (b) {
		if (b & 1) r = 1ll * r * w % p;
		w = 1ll * w * w % p;
		b >>= 1;
	}
	return r;
}

int cal(int x) {
	int ret = 1, ci;
	for (int i = 1, pi = 2; i <= num && pi * pi <= x; pi = prime[++i]) {
		if (x % pi == 0) {
			ci = 1; x /= pi;
			while (x % pi == 0) x /= pi, ++ci;
			ret = 1ll * ret * (1ll * (ipow(pi, 2 * ci + 1) - pi + p) % p * ni[pi + 1] % p + 1) % p;
		}
	}
	if (x > 1) {
		ret = 1ll * ret * ((1ll * x * x % p * x % p - x + p) % p * ipow(x + 1, p - 2) % p + 1) % p;
	}
	return ret;
}

int main() {
	Euler_shai();
	
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		printf("%lld
", (1ll * n * ni[2] % p * (cal(n) - 1) % p + n) % p);
	}
	return 0;
}