Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3587 Accepted Submission(s): 1419
Problem Description
Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.
Input
The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 1015) and (1 <=N <= 109).
Output
For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.
Sample Input
2
1 10 2
3 15 5
Sample Output
Case #1: 5
Case #2: 10
Hint
In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.
#include<stdio.h> #include<string.h> using namespace std; long long a[1000],num; void prime(long long n) { long long i; num=0; for(i=2;i*i<=n;i++) { if(n%i==0) { a[num++]=i; while(n%i==0) n=n/i; } } if(n>1) a[num++]=n; } long long get(long long m) { long long que[10000],k,t=0,sum=0; que[t++]=-1; for(int i=0;i<num;i++) { k=t; for(int j=0;j<k;j++) que[t++]=que[j]*a[i]*(-1); } for(int i=1;i<t;i++) sum=sum+m/que[i]; return sum; } int main() { long long T,x,y,n,cnt; while(scanf("%lld",&T)!=EOF) { for(int i=1;i<=T;i++) { scanf("%lld%lld%lld",&x,&y,&n); prime(n); cnt=y-get(y)-(x-1-get(x-1)); printf("Case #%d: ",i); printf("%lld\n",cnt); } } return 0; }