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
#include<cstdio>#include<cstring>using namespace std;#define MAXN 200000#define lint __int64struct 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); }}#include<cstdio>#include<cstring>using namespace std;#define MAXN 200000#define lint __int64struct 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); }}
