给定正整数,求 ∑ i = 1 n g c d ( i , n ) \sum_{i=1}^n{gcd(i,n)} ∑i=1ngcd(i,n) n<=2^32
枚举gcd: ∑ i = 1 n g c d ( i , n ) = ∑ d ∣ n d ∑ i = 1 n [ g c d ( i , n ) = d ] \sum_{i=1}^ngcd(i,n)=\sum_{d|n}d\sum_{i=1}^n[gcd(i,n)=d] ∑i=1ngcd(i,n)=∑d∣nd∑i=1n[gcd(i,n)=d] = ∑ d ∣ n d ∑ i = 1 n / d [ g c d ( i , n / d ) = 1 ] \sum_{d|n}d\sum_{i=1}^{n/d}[gcd(i,n/d)=1] ∑d∣nd∑i=1n/d[gcd(i,n/d)=1] = ∑ d ∣ n d ∗ p h i ( n / d ) \sum_{d|n}d*phi(n/d) ∑d∣nd∗phi(n/d)
枚举n的约数可以直接求