UVa 11426

《训练指南》p.125

设f[n] = gcd(1, n) + gcd(2, n) + …… + gcd(n - 1, n);

则所求答案为S[n] = f[2]+f[3]+……+f[n];

求出f[n]即可递推求得S[n]:S[n] = S[n - 1] + f[n];

所有gcd(x, n)的值都是n的约数，按照约数进行分类，令g(n, i)表示满足gcd(x, n) = i && x < n 的正整数x的个数，则f[n] = sum{ i * g(n, i) | n % i = 0 };

gcd( x, n ) = i 的充要条件为：gcd( x / i, n / i ) = 1; 因此满足条件的x/i有phi(n/i)个，说明g(n, i) = phi( n/i )；

如果依次计算f[n]，枚举f[n]的约数的话效率太低

因此对于每个i枚举它的倍数n并更新f[n]，时间复杂度与素数筛法同阶。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

#define LL long long int

using namespace std;

const int MAXN = 4000100;

LL phi[MAXN];
LL S[MAXN];
LL f[MAXN];

//筛法计算欧拉数
void phi_table( int n )
{
    for ( int i = 2; i < n; ++i ) phi[i] = 0;
    phi[1] = 1;
    for ( int i = 2; i < n; ++i )
        if ( !phi[i] )
        {
            for ( int j = i; j < n; j += i )
            {
                if ( !phi[j] )
                    phi[j] = j;
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    return;
}

int main()
{
    phi_table( MAXN );

    memset( f, 0, sizeof(f) );
    for ( int i = 1; i < MAXN; ++i )
        for ( int j = i * 2; j < MAXN; j += i )
            f[j] += i * phi[j / i];

    S[2] = f[2];
    for ( int i = 3; i < MAXN; ++i )
        S[i] = S[ i - 1 ] + f[i];

    int n;
    while ( scanf( "%d", &n ), n )
    {
        printf("%lld
", S[n] );
    }
    return 0;
}