hdu4135 Co-prime_hdu4135frankax-优快云博客

Co-prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6933 Accepted Submission(s): 2733

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}.

代码：

#include<cstdio>
#include<algorithm>
#include<map>
#define ll long long
using namespace std;
map<ll,ll> mp;
ll a[2000];
int pos;
void solve(ll n)
{
   pos = 0;
   for (ll i = 2;i * i <= n;i ++)
   {
       if (n % i == 0)
       {
           while (n % i == 0)
           {
               a[pos] = i;
               n /= i;
           }
           pos ++;
       }
   }
   if (n != 1) a[pos ++] = n;
}
ll exc(ll n,ll m)
{
   ll ans = 0;
   for (int i = 1;i < (1 << m);i ++)
   {
       int cnt = 0,lm = 1;
       for (int j = 0;j < m;j ++)
       {
           if ((i >> j) & 1)
           {
               cnt += 1;
               lm *= a[j];
           }
       }
       if (cnt & 1) ans += n / lm;
       else ans -= n / lm;
   }
   return ans;
}
int main()
{
   int t,kase = 0;
   scanf("%d",&t);
   while (t--)
   {
       kase ++;
       ll n,a,b;
       scanf("%lld %lld %lld",&a,&b,&n);
       solve(n);
       ll numb = exc(b,pos);
       ll numa = exc(a - 1,pos);
       printf("Case #%d: %lld\n",kase,b - a + 1 - (numb - numa));
   }
   return 0;
}