一道數學題

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求 ∑ni=1∑ij=1gcd(i,j) 的值。

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gcd(i∗d,j∗d)=d∗gcd(i,j)

設 gcd(i,j)=1 ， 不大於 i 且與 i 互質的數的個數是 φ(i)

所以 ∑ni=1∑ij=1gcd(i,j)=∑nd=1∑ndi=1φ(i)∗d=∑nd=1d∗∑ndi=1φ(i)

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枚舉 d ，預處理求出 φ(i) 的前綴和即可。

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#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <utility>
#include <iostream>
#include <algorithm>

template<class Num>void read(Num &x)
{
    char c; int flag = 1;
    while((c = getchar()) < '0' || c > '9')
        if(c == '-') flag *= -1;
    x = c - '0';
    while((c = getchar()) >= '0' && c <= '9')
        x = (x<<3) + (x<<1) + (c-'0');
    x *= flag;
    return;
}
template<class Num>void write(Num x)
{
    if(x < 0) putchar('-'), x = -x;
    static char s[20];int sl = 0;
    while(x) s[sl++] = x%10 + '0',x /= 10;
    if(!sl) {putchar('0');return;}
    while(sl) putchar(s[--sl]);
}
#define REP(__i,__start,__end) for(int __i = (__start); __i <= (__end); __i++)

const int size = 1e7 + 50;

int n, phi[size];
long long sum[size];
int prime[size], tot;
bool check[size];

void prework()
{
    sum[1] = phi[1] = 1;

    REP(i, 2, n)
    {
        if(!check[i])
        {
            prime[++tot] = i;
            phi[i] = i - 1;
        }

        REP(j, 1, tot)
        {
            if((long long)i * prime[j] > n) break;

            check[i * prime[j]] = true;

            if(i % prime[j])
            {
                phi[i * prime[j]] = phi[i] * phi[prime[j]];
            }
            else
            {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
        }

        sum[i] = sum[i - 1] + phi[i];
    }

}

void solve()
{
    long long ans = 0;

    REP(i, 1, n) ans += sum[n / i] * i;

    write(ans);
}

int main()
{
    freopen("A.in","r",stdin);
    freopen("A.out","w",stdout);

    read(n);

    prework();

    solve();

    fclose(stdin);
    fclose(stdout);
    return 0;
}