给定正整数N,求LCM(1,N)+LCM(2,N)+...+LCM(N,N)。
多组询问,1≤T≤300000,1≤N≤1000000
(sum_{i=1}^nlcm(i,n))
(=sum_{i=1}^nfrac{in}{gcd(i,n)})
(=nsum_{p|n}frac 1 psum_{i=1}^ni[gcd(i,n)=p])
(=nsum_{p|n}sum_{i=1}^{n/p}i[gcd(i,frac n p)=1])//注意这里是n/p而不是n,mdzz第一次我这里推错了
(=nsum_{p|n}sum_{i=1}^{p}i[gcd(i,p)=1])
然后就可以套用公式(sum_{i=1}^ni[gcd(i,n)=1]=frac {[n=1]+nvarphi(n)}{2})
(=nsum_{p|n}frac{[p=1]+pvarphi(p)}{2})
线性筛,对于所有(p)预处理所有的([p=1]+varphi(p)*p)
然后枚举倍数,在100Wlog100W时间复杂度内求出所有N的答案
代码没时间写了,明天放
然而还是在3分钟内写出了代码
#include <cstdio>
using namespace std;
const int fuck = 1000000;
bool vis[fuck + 10];
int prime[fuck + 10], tot;
int phi[fuck + 10];
long long ans[fuck + 10];
int main()
{
phi[1] = 1;
for (int i = 2; i <= fuck; i++)
{
if (vis[i] == false) prime[++tot] = i, phi[i] = i - 1;
for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
{
vis[i * prime[j]] = true;
if (i % prime[j] == 0) { phi[i * prime[j]] = phi[i] * prime[j]; break; }
else phi[i * prime[j]] = phi[i] * (prime[j] - 1);
}
}
for (int p = 1; p <= fuck; p++)
{
long long sb = (p == 1) + phi[p] * (long long)p;
for (int b = p; b <= fuck; b += p)
{
ans[b] += sb;
}
}
int t; scanf("%d", &t);
while (t --> 0)
{
int x;
scanf("%d", &x);
printf("%lld
", ans[x] * x / 2);
}
return 0;
}