为什么∑i=1n∑j=i+1ngcd(i,j)=∑i=1nf(i)\sum\limits^n_{i=1}\sum\limits^n_{j=i+1}\gcd(i,j) =\sum\limits^n_{i=1}f(i)i=1∑nj=i+1∑ngcd(i,j)=i=1∑nf(i)
