LightOj 1375(歐拉變形)

lightoj 1375
題目大意：
求小於n 的數字中，任意兩個不相等的數字的lcm 和，即∑i=1,j=1,i≠ji=n,j=nlcm(i,j) ;
思路：
lcm(1,2)  lcm(1,3)  lcm(1,4)  … lcm(1,n)
                  lcm(2,3)  lcm(2,4)  … lcm(2,n)
                                   lcm(3,4)  … lcm(3,n)
                                                     …lcm(n−1,n)
即可求∑i=2n∑j=1j=i−1lcm(j,i) ;
lcm(i,n)=i∗ngcd(i,n) , 令gcd(i,n)=G;
lcm(i,n)=ig∗ng∗g;  gcd(ig,ng)=1 ;
又已知小於n 且與n 互質的數的和等於n∗φ(n)2 , 證明;
∑i=1n−1lcm(i,n)=ig∗φ(ng)2∗g ;
即: ∑i=1n−1lcm(i,n)=∑d∣n,d≠nd∗φ(d)2∗n ; 即枚舉n的因子；

取模264 , 變量定義爲unsigned  long  long ; 溢出264 即爲取模264 ;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#define N 3000001
#define LL unsigned long long

using namespace std;

int euler[N];
LL f[N], a[N];

void init_Euler()
{
    euler[1] = 1;

    for (int i = 2; i < N; i++)
    {
        euler[i] = i;
    }

    for (int i = 2; i < N; i++)
    {
        if (euler[i] == i)
        {
            for (int j = i; j < N; j += i)
            {
                euler[j] = euler[j] / i * (i - 1);
            }
        }
    }

    memset(a, 0, sizeof(a));  
    memset(f, 0, sizeof(f)); 

    for(int i = 2; i < N; i++)  
    {  
        for(int j = i; j < N; j += i)  
        {  
            f[j] += ((LL)euler[i]) * i / 2 * j;  
        }  

        a[i] = a[i - 1] + f[i];  
    }
}

int main()
{
    init_Euler();

    int T;
    scanf("%d", &T);

    for (int cas = 1; cas <= T; cas++)
    {
        int n;
        scanf("%d", &n);

        printf("Case %d: %llu\n", cas, a[n]);
    }
    return 0;
}