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【NOI2010】能量采集 题解
谨纪念我的第一道手推出来的莫反题。
题目大意:已知 n n n, m m m,求 ∑ i = 1 n ∑ j = 1 m ( 2 ⋅ gcd ( i , j ) − 1 ) \sum\limits_{i=1}^n\sum\limits_{j=1}^m(2\cdot \gcd(i,j)-1) i=1∑nj=1∑m(2⋅gcd(i,j)−1)。
首先变形一手:
∑ i = 1 n ∑ j = 1 m ( 2 ⋅ gcd ( i , j ) − 1 ) = 2 ∑ i = 1 n ∑ j = 1 m gcd ( i , j ) − n × m \sum\limits_{i=1}^n\sum\limits_{j=1}^m(2\cdot\gcd(i,j)-1)=2\sum\limits_{i=1}^n\sum\limits_{j=1}^m\gcd(i,j)-n\times m i=1∑nj=1∑m(2⋅gcd(i,j)−1)=2i=1∑nj=1∑mgcd(i,j)−n×m
然后我们只用求出中间那两个 ∑ \sum ∑ 就好了。
∑ i = 1 n ∑ j = 1 m gcd ( i , j ) = ∑ i = 1 n ∑ j = 1 m ∑ d = 1 n d [ gcd ( i , j ) = d ] = ∑ d = 1 n d ∑ i = 1 ⌊ n d ⌋ ∑ j = 1 ⌊ m d ⌋ [ gcd ( i , j ) = 1 ] = ∑ d = 1 n d ∑ i = 1 ⌊ n d ⌋ ∑ j = 1 ⌊ m d ⌋ ∑ x ∣ gcd ( i , j ) μ ( x ) = ∑ d = 1 n d ∑ x = 1 ⌊ n d ⌋ μ ( x ) ⌊ n d x