Hdu4135 Co-prime

Co-prime

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 49   Accepted Submission(s) : 21

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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 <= 10 ¹⁵) and (1 <=N <= 10 ⁹).

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}.

Source

The Third Lebanese Collegiate Programming Contest

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题目的意思是给出一段区间求与给出的k互质的数有几个

思路：先把问题转化为前缀和问题，然后算出与k有公因数有多少个，可以先对k分解质

因 数，这样的数的种类肯定不多

然后dfs求每种数的个数，用容斥原理计数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>

using namespace std;

#define LL long long
const int inf=0x7fffffff;

LL p[100005];
LL ans,l,r;
int tot;

void dfs(int pos,LL x,int cnt)
{
    if(pos>=tot)
    {
        if(cnt==0)
            return;
        if(cnt%2)
        {
            ans+=r/x;
            ans-=(l-1)/x;
        }
        else
        {
            ans-=r/x;
            ans+=(l-1)/x;
        }
        return;
    }
    dfs(pos+1,x*p[pos],cnt+1);
    dfs(pos+1,x,cnt);
}

int main()
{

  int T;
  LL k;
  scanf("%d",&T);
  int q=1;
  while(T--)
  {
      tot=0;
      scanf("%lld%lld%lld",&l,&r,&k);
     for(LL i=2;i*i<=k;i++)
     {
         if(k%i==0)
            p[tot++]=i;
         while(k%i==0) k/=i;
     }
     if(k>1) p[tot++]=k;
     ans=0;
    dfs(0,1,0);
    ans=r-l+1-ans;
    printf("Case #%d: %lld\n",q++,ans);
  }
    return 0;
}