Given you a number n, find the number of coprime in the interval [a, b] and N.

Problem-solving ideas: To find the number of [a, b] interval and n coprime, we can convert to [1,b] the number of n coprime minus the number of [1,a-1] and N coprime

and the number of n coprime required in [1,b], we can naturally think of the principle of repulsion

The number of coprime and K in the solution interval [a, b]
First, the K is decomposed by prime number, and then with this factor gcd!=1 a total of n/ki

There is a repetition of the calculation, so the use of the principle of repulsion: K1+k2+k3- k1*k2-..... Find the number with K not coprime, with total minus can

Description given you a number n, please count the number of the numbers in the interval and n coprime in [A, b]. Two number coprime when and only if they have no other public factors other than 1 or their largest public factor is 1. 1 and any number are mutually-voxel. Input the first line to enter an integer t (1 <= T <= 100), representing the T Group test data. Next T line, 3 integers per line a,b,n (1 <= a <=b <=10^15, 1<= n <= 10^9), separated by a space. Output outputs an integer representingand NCoprimenumber of the numbers. Sample Input

21 10 23) 10 5

Sample Output

56

<pre name= "code" class= "CPP" > #include <stdio.h> #include <iostream>#       Include<string.h>using namespace Std;long long s[1020],k;void prime (long long m)//For a number of prime factor {long long I;    k=0;            for (i=2;i*i<=m;i++) {if (m%i==0) {s[k++]=i;        while (m%i==0) m/=i;//printf ("%d\n", m);            }} if (m>1) s[k++]=m; printf ("###\n");}    Long Long Quc (long long m)//queue array implements the tolerant principle {long long p[10020],i,j,t=0,sum=0,z;    P[t++]=-1;        for (i=0;i<k;i++) {z=t;        for (j=0;j<z;j++) {p[t++]=p[j]*s[i]* (-1);    }} for (i=1;i<t;i++) sum+=m/p[i]; return sum;}    int main () {long long n;    scanf ("%lld", &n);        while (n--) {Long long a,b,m,sum;           scanf ("%lld%lld%lld", &a,&b,&m);        Prime (m);        Sum=b-quc (b)-(A-1) +quc (A-1);    printf ("%lld\n", sum); }}

Given you a number n, find the number of coprime in the interval [a, b] and N.