HDU 3501 Calculation 2 (Euler function application)

Calculation 2 Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total Submission (s): 2989 Accepted Submission (s): 1234

problem DescriptionGiven A positive integer n, your task is to calculate the sum of the positive integers less than N which was not copri Me to N. A is said to being coprime to B if A, B share no common positive divisors except 1. Inputfor each test case, the There is a line containing a positive integer N (1≤n≤1000000000). A line containing a single 0 follows the last test case. Outputfor each test case, you should print the sum of module 1000000007 in a line. Sample Input

340

 Sample Output

02

 AuthorGtmac Sourceacm-icpc multi-university Training Contest (7)--host by hits

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=3501

The main idea: to find less than N and n non-coprime number of

Title Analysis: Considering the number of n coprime and is less than N and very good, and then subtract with the sum, less than n and n coprime number and equal to N*phi[n]/2,phi[n] is n corresponding Euler function value, proved as follows:
Set GCD (n, i) = = 1, then have GCD (n, n-i) = = 1
This can be disproved: suppose there is a k! = 1,GCD (n, n-i) = = k, then there is n%k = = 0 and (n-i)% K = = 0 that is (n k-i% k)% K = = 0, get (i% k)% K = = 0
So I must be a multiple of k, so GCD (n, i) = = k, which conflicts with GCD (n, i) = = 1, so for GCD (n, i) = = 1, there is GCD (n, n-i) = = 1, so the two logarithm with n coprime is n and the total number of n coprime is phi[n], note here does not appear n = = 2*i case because 2,phi[i] is even
So the sum of the number of n coprime is less than n and equal to N*phi[n]/2, so the final answer is (n * (n-1)/2-n*phi[n]/2)% MOD, which asks the complexity of the single Phi to be sqrt ((n)

#include <cstdio> #include <cmath> #define LL long longint Const MOD = 1e9 + 7;int phi (int x) {    int res = x;< c1/>for (int i = 2; I * i <= x; i++)    {        if (x% i = = 0)         {            res = res/i * (i-1);            while (x% i = = 0)                 x/= i;        }    }    if (x > 1)         res = res/x * (x-1);    return res;} int main () {    int n;    while (scanf ("%d", &n)! = EOF && N)    {        ll sum = (LL) n * (n-1)/2;        ll copsum = (LL) n * PHI (n)/2;        ll ans = sum-copsum;        printf ("%i64d\n", ans% MOD);}    }

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HDU 3501 Calculation 2 (Euler function application)