POJ 2480 Longge's problem 欧拉函数

公式：f(N)=∑x*φ(N/x)，x | N （x是N的约数）
因为在1···N中，gcd(i，N) = x， 的个数的等于φ(N / x)

另外还可以利用函数的积性：

对于正整数n的一个函数 f(n)，当中f(1)=1且当a,b互质，f(ab)=f(a)f(b)，在数论上就称它为积性函数。若某函数f(n)符合f(1)=1，且就算a,b不互质，f(ab)=f(a)f(b)，则称它为完全积性函数。

不妨令M, N互素
f(M) = ∑d1 * φ(M / d1), d1 | M
f(N) = ∑d2 * φ(N / d2), d2 | N

f(MN) = ∑d * φ(MN / d), d | MN
因为M, N互素，则每个d都可以唯一分解为M中的因子d1, 和N中的因子d2
即d = d1 * d2, d1 | M, d2 | N, d1与d2互素
则d * φ(MN / d) = d1 * d2 * φ(M / d1) * φ(N / d2)
f(MN)中的项与f(M) * f(N)中的项一一对应

解法一：47MS

#include<cstdio>
#include<cstring>
using namespace std;#define MAXN 200000
#define lint __int64
struct Factor { lint b, e; };
Factor f[MAXN]; lint fnum;
lint a[MAXN], p[MAXN], pn;
lint n, ret;void Prime()
{lint i, j; pn = 0;memset(a,0,sizeof(a));for ( i = 2; i < MAXN; i++ ){if ( a[i] == 0 ) p[pn++] = i;for ( j = 0; j < pn && i*p[j] < MAXN && (p[j]<=a[i] || !a[i]); j++ )a[i*p[j]] = p[j];}
}lint Euler ( lint n )
{lint ret = n;for ( int i = 0; p[i] * p[i] <= n; i++ ){if ( n % p[i] == 0 ){ret = ret - ret / p[i];while ( n % p[i] == 0 ) n /= p[i];}}if ( n > 1 )ret = ret - ret / n;return ret;
}void split ( lint n )
{fnum = 0;for ( int i = 0; p[i] * p[i] <= n; i++ ){if ( n % p[i] ) continue;f[fnum].b = p[i]; f[fnum].e = 0;while ( n % p[i] == 0 ){f[fnum].e++;n /= p[i];}fnum++;}if ( n > 1 )f[fnum].b = n, f[fnum++].e = 1;}void DFS ( lint val, int index ) //求n的每一个约数，然后利用欧拉函数
{if ( index == fnum ){ret += Euler(n/val) * val;   //Euler(n/val)的值表示1-n中gcd(n,i）= val的个数return;}for ( lint i = 0, tmp = 1; i <= f[index].e; i++, tmp *= f[index].b )DFS ( val*tmp, index+1 );
}int main()
{Prime();while ( scanf("%I64d",&n) != EOF ){split ( n );ret = 0;DFS ( 1, 0 );printf("%I64d\n",ret);}
}

解法二：利用积性16ms

#include<cstdio>
#include<cstring>
using namespace std;#define MAXN 200000
#define lint __int64
struct Factor { lint b, e, mult; };
Factor f[MAXN]; lint fnum;
lint a[MAXN], p[MAXN], pn;void Prime()
{lint i, j; pn = 0;memset(a,0,sizeof(a));for ( i = 2; i < MAXN; i++ ){if ( a[i] == 0 ) p[pn++] = i;for ( j = 0; j < pn && i*p[j] < MAXN && (p[j]<=a[i] || !a[i]); j++ )a[i*p[j]] = p[j];}
}lint Euler ( lint n )
{lint ret = n;for ( int i = 0; p[i] * p[i] <= n; i++ ){if ( n % p[i] == 0 ){ret = ret - ret / p[i];while ( n % p[i] == 0 ) n /= p[i];}}if ( n > 1 )ret = ret - ret / n;return ret;
}void split ( lint n )
{fnum = 0;for ( int i = 0; p[i] * p[i] <= n; i++ ){if ( n % p[i] ) continue;f[fnum].b = p[i]; f[fnum].e = 0;f[fnum].mult = 1;while ( n % p[i] == 0 ){f[fnum].e++;f[fnum].mult *= p[i];n /= p[i];}fnum++;}if ( n > 1 )f[fnum].b = f[fnum].mult = n, f[fnum++].e = 1;}int main()
{Prime(); lint n;while ( scanf("%I64d",&n) != EOF ){split ( n );lint ret = 1, tmp, sum;for ( int i = 0; i < fnum; i++ ){tmp = 1, sum = Euler(f[i].mult); //所有与f[i].mult互素的数先加起来for ( int j = 1; j <= f[i].e; j++ ){tmp *= f[i].b;sum += Euler(f[i].mult/tmp) * tmp;}ret *= sum;}printf("%I64d\n",ret);}
}