Bzoj 2005: [Noi2010] Energy harvesting

1#include <cstdio>2#include <iostream>3 #definell Long Long4 using namespacestd;5ll f[100009],n,m,ans;6 intMain ()7 {8scanf"%d%d",&n,&m);9     if(n>m)Ten swap (n,m); One      for(inti=n;i;i--) A       { -f[i]= (n/i) * (m/i); -          for(intj=2; j*i<=n;j++) thef[i]-=f[i*j]; -ans+=f[i]* (2*i-1); -       } -printf"%lld\n", ans); +     return 0; -}

There is a conclusion that the number of endpoints on the (x, y) line is equal to GCD (x, y), prove that they are yy, then the problem becomes GCD and, set F[i] for the GCD for I of the number of points, the method can be nlogn in the program to find out.

Bzoj 2005: [Noi2010] Energy harvesting