spoj LCM Sum

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一个数论定理对于n >= 2 小于它与它互质的数的和为 n*phi*(n)/2
然后lcm（a,b） = a * b / gcd(a, b)
时间太紧所以打表预处理枚举每个数然后找到每个能整除它的数加上一部分欧拉函数的和

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<iostream>#include<cmath>#define maxn 1000005#define LL long longusing namespace std;// 欧拉函数3LL eular[maxn];LL prime[maxn][15],num[maxn],flag[maxn],AM = 0;void Eorue(){    eular[1] = 1;    for(LL i = 2;i < maxn; i++){        if(eular[i] == i){            eular[i] = i - 1;            for(LL j = 2; j * i < maxn; j++){                eular[i * j] = eular[i * j] * (i - 1) / i;            }        }        //eular[i] += eular[i-1];    }}LL ans[maxn];int main(){    for(LL i = 1; i < maxn; i++) eular[i] = i;    Eorue();    for(LL i = 1; i <= 1000000; i++)        for(LL j = 1; j * i <= 1000000; j++)         {            if(j != 1)ans[j * i] += j * i * (eular[j] * j / 2ll);            else ans[j * i] += j * i;         }    int t;    scanf("%d", &t);    while(t--)    {        int n;        scanf("%d", &n);        printf("%lld\n", ans[n]);    }    return 0;}//277811686426000000

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