UVa 10214 (Möbius inversion or Euler function) Trees in a Wood.

Test instructions

This question and POJ 3090 very similar, asks |x|≤a,|y|≤b to stand in the origin point to see the whole hour the number k, all the hour number is N (removes the origin), asks k/n

Analysis:

There are four visible points on the axis because the number of points visible in each quadrant is the same, so we only ask for the number of points visible in the first quadrant and then X4+4, which is K.

The visible point satisfies gcd (x, y) = 1, so the problem is converted to x∈[1, a], y∈[1, b], and the number of gcd (x, y) = 1.

Analogy HDU 1695 can be used Möbius inversion to do, I also wrote the normal and block acceleration of the two code, turn to find the running time difference is not too much.

1#include <cstdio>2#include <algorithm>3typedefLong LongLL;4 5 Const intMAXN = -;6 intmu[maxn+Ten], vis[maxn+Ten], PRIME[MAXN];7 8 voidMobius ()9 {Tenmu[1] =1; One     intCNT =0; A      for(inti =2; I <= MAXN; ++i) -     { -         if(!Vis[i]) the         { -Mu[i] =-1; -prime[cnt++] =i; -         } +          for(intj =0; J < CNT && (LL) i*prime[j] <= maxn; ++j) -         { +VIS[I*PRIME[J]] =1; A             if(i% prime[j]! =0) Mu[i*prime[j]] =-Mu[i]; at             Else -             { -MU[I*PRIME[J]] =0; -                  Break; -             } -         } in     } -     //compute prefixes and, for block acceleration to      for(inti =2; I <= -; ++i) Mu[i] + = mu[i-1]; +  - } the  * intMain () $ {Panax Notoginseng Mobius (); -     intA, B; the      while(SCANF ("%d%d", &a, &b) = =2) +     { A         if(A = =0&& b = =0) Break; theLL K =0, N = (LL) (A *2+1) * (b*2+1) -1; +         if(A >b) Std::swap (A, b); -          for(inti =1, J; I <= A; i = j +1) $         { $j = Std::min (A/(a/i), b/i)); -K + = (LL) (Mu[j]-mu[i-1]) * (a/i) * (b/i); -         } the         //for (int i = 1; i <= A; ++i) K + = (LL) mu[i] * (a/i) * (b/i); -K = (+ K1)*4;Wuyi  theprintf"%.7f\n", (Double* H/N); -     } Wu  -     return 0; About}

code June

You can also follow the purple book on the idea of Euler functions, because a has a smaller range, so you can count by column. But this method is much slower than the Möbius inversion.

1#include <cstdio>2#include <cmath>3typedefLong LongLL;4 5 intPhiintN)6 {7     intm = sqrt (n +0.5);8     intAns =N;9      for(inti =2; I <= m; ++i)if(n% i = =0)Ten     { OneAns = ans/i * (i-1); A          while(n% i = =0) n/=i; -     } -     if(N >1) ans = ans/n * (n1); the     returnans; - } -  - intgcdintAintb) + { -     returnb = =0? A:GCD (b, a%b); + } A  at intMain () - { -     intA, B; -      while(SCANF ("%d%d", &a, &b) = =2&&a) -     { -ll N = (ll) (A *2+1) * (b*2+1) -1; inLL K =0; -          for(intx =1; x <= A; ++x) to         { +             intK = b/x; -K + = Phi (x) *K; the              for(inty = k*x+1; Y <= b; y++) *                 if(GCD (x, y) = =1) k++; $         }Panax NotoginsengK = (+ K1)*4; -printf"%.7f\n", (DoubleKN); the     } +  A     return 0; the}

code June two

UVa 10214 (Möbius inversion or Euler function) Trees in a Wood.