51nod 1040 Greatest common divisor and (Euler function)

1040 Greatest Common Divisor Source: rihkddd Base time limit: 1 seconds space limit: 131072 KB score: 80 Difficulty: 5 level algorithm problem collection attention to give a N, ask 1-n this n number, with the greatest common divisor of N. For example: n = 61,2,3,4,5,6 the same as 6 greatest common divisor, respectively, is 1,2,3,2,1,6, add together = Input

1 number N (n <= 10^9)

Output

Sum of the number of conventions

Input example

6

Output example

15

Idea: The aim is to seek ∑ (i= 1,n) gcd (i, n); gcd (i, N) = x, which means that x is a factor of N. Slightly deformed gcd (i/x, n/x) = 1, see this equation can think of Euler function, that is, to find a smaller than n/x and its coprime number. Because these books and n/x coprime, multiply X and N greatest common divisor only X. That is, we first find out each factor, and then calculate how much each factor contributes.

/** author:sweat123 * Created time:2016/6/27 14:01:46 * File Name:main.cpp*/#include<Set>#include<map>#include<queue>#include<stack>#include<cmath>#include<string>#include<vector>#include<cstdio>#include<time.h>#include<cstring>#include<iostream>#include<algorithm>#defineINF 1<<30#defineMOD 1000000007#definell Long Long#defineLson l,m,rt<<1#defineRson m+1,r,rt<<1|1#definePi ACOs (-1.0)using namespacestd;Const intMAXN =1000000;intn;ll EF (intN) {ll cnt=N; inti;  for(i =2; I * I <= N; i++){         if(n% i = =0) {CNT-= cnt/i;//(X-X/P1) * (1-1/P2) * (1-1/P3) * (1-1/P4) .....             while(n% i = =0) n/=i; }     }    if(N >1) CNT-= CNT/N; returnCNT;} intMain () { while(~SCANF ("%d",&N)) {ll ans=0;  for(inti =1; I <= (int) sqrt (n); i++){            if(n% i = =0) {ans+ = EF (n/i) *i; if(n/i! =i) {ans+ = EF (i) * (N/i); }}} printf ("%lld\n", ans); }    return 0;}

51nod 1040 Greatest common divisor and (Euler function)