Hdu4135--co-prime (Euler function + repulsion principle)

Source: Internet
Author: User
Tags greatest common divisor

Co-PrimeTime limit:MS Memory Limit:32768KB 64bit IO Format:%i64d &%i64u SubmitStatusAppoint Description:System Crawler (2015-01-07)

Description

Given A number N, you is asked to count the number of integers between a and B inclusive which is relatively prime to N.
Integers is said to be co-prime or relatively prime if they has no common positive divisors other than 1 or, equival Ently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

Input

The first line in input contains T (0 < T <=) The number of test cases, each of the next T lines contains three Integers a, B, N where (1 <= A <= B <=) and (1 <=n <= 10 9).

Output

For each test case, print the number of integers between A and B inclusive which is relatively prime to N. follow the Out Put format below.

Sample Input

Sample Output

Hint

In the first test case, the five integers in range [1,10] which is relatively prime to 2 is {1,3,5,7,9}.          


Calculate the number of coprime in a to B, and N

Statistics 1 to A-1, and 1 to B and N coprime number, in subtraction, 1 to M and N coprime number, first 1 to M and n not coprime number.

To find the number of mass after n decomposition, with a binary number to denote the selection and non-selection of the first prime, to get the number contained in M, the election of odd number of prime numbers is, the results of the statistical summation, for even number, minus.


#include <cstdio> #include <cstring> #include <algorithm>using namespace std; #define LL __int64int   PRIM[1000000], vis[1000000], cnt, void sieve () {memset (vis,0,sizeof (VIS));   CNT = 0;   LL I, J;           for (i = 2; I <= 1000000; i++) {if (!vis[i]) {prim[cnt++] = i;       for (j = i*i; J < 1000000; J + = i) vis[j] = 1; }}}ll p[1000], p_num;    ll F (ll N,ll m) {ll k = n, temp, ans = 0;    int I, j, num;        for (i = 0, p_num = 0; i < cnt; i++) {if (k% prim[i] = = 0) {p[p_num++] = Prim[i];        } while (k% prim[i] = = 0) {k/= prim[i];    } if (k = = 1) break;    } if (K > 1) p[p_num++] = k;            for (i = 1; i < (1<<p_num); i++) {for (j = 0, num = 0, temp = 1; j < P_num; J + +) {           if ((1<<j) & i) {temp *= p[j];     num++;        }} if (num% 2) ans + = m/temp;    else ans-= m/temp; } return ans;    int main () {LL T, TT, A, b, N;    Sieve ();    scanf ("%i64d", &t);        for (TT = 1; TT <= t; tt++) {scanf ("%i64d%i64d%i64d", &a, &b, &n);    printf ("Case #%i64d:%i64d\n", TT, (B-f (N,b))-(A-1-f (N,a-1))); } return 0;}


Hdu4135--co-prime (Euler function + repulsion principle)

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