hdoj Co-prime 4135 （容斥原理）

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2778    Accepted Submission(s): 1070

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

21 10 23 15 5

Sample Output

Case #1: 5Case #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}. 
 
思路：
要求在l--r这一范围内与n互质的数的个数，先求出在此范围内不与n互质的数的个数num，然后用总个数减去num即为所求。
应用容斥原理解。
p数组用来存放n的因子，q数组用来存放各个因子的乘积。
#include<stdio.h>#include<string.h>#define ll long longll p[10010];ll q[10010];ll k;void getp(ll n){k=0;ll i,j;for(i=2;i*i<=n;i++){if(n%i==0){p[k++]=i;while(n%i==0)n/=i;}}if(n>1)p[k++]=n;}ll solve(ll n){ll i,j,t=0,kk,sum=0;q[t++]=-1;for(i=0;i<k;i++){kk=t;for(j=0;j<kk;j++)q[t++]=p[i]*q[j]*-1;}for(i=1;i<t;i++)sum+=(n/q[i]);return sum;}int main(){int t,T=1;ll l,r,n;scanf("%d",&t);while(t--){scanf("%lld%lld%lld",&l,&r,&n);getp(n);ll cnt=(r-solve(r))-(l-1-solve(l-1));printf("Case #%d: %lld\n",T++,cnt);}return 0;}