hdu4135 Co-prime

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 87    Accepted Submission(s): 39

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

 

Recommend

lcy

一看就是容斥原理的题，提取质因子然后容斥就可以了，初始化wa了一次，好久没做题了总犯些弱智错误……

代码：

#include <stdio.h>
#include <math.h>

int p[100];
int up;
long long ansa,ansb,ans;
long long a,b;

void DFS(int n,bool tag,long long num)
{
    if (n==up)
    {
        if (tag==1)
        {
            ansa-=a/num;
            ansb-=b/num;
        }
        else
        {
            ansa+=a/num;
            ansb+=b/num;
        }
        return;
    }
    DFS(n+1,tag,num);
    DFS(n+1,!tag,num*p[n]);
}

int main()
{
    int i,j,n,T,k,cnt;
    cnt=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%d",&a,&b,&n);
        a--;
        ansa=ansb=0;
        up=0;
        k=n;
        for (i=2;i<=sqrt(n*1.0);i++)
        {
            if (k%i==0)
            {
                while(k%i==0) k=k/i;
                p[up++]=i;
            }
        }
        if (k!=1)
        {
            p[up++]=k;
        }
        DFS(0,0,1);
      //  printf("%lld..%lld\n",ansb,ansa);
        printf("Case #%d: %I64d\n",cnt++,ansb-ansa);
    }
    return 0;
}