Co-prime(hdu4135)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3313    Accepted Submission(s): 1286

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}.

 

Source

The Third Lebanese Collegiate Programming Contest

 

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思路：素数打表+容斥原理;

因为要求在[n，m]中与互质的数的个数。

先打表求素数，然后分解k，求出k由哪些素数组成，然后我们可以用容斥求出[n,m]中与k不互质的数，然后区间长度减下即可；

每个数的质因数个数不会超过20个。

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<stdlib.h>
  5 #include<string.h>
  6 #include<vector>
  7 #include<queue>
  8 #include<stack>
  9 using namespace std;
 10 long long  gcd(long long n,long long  m);
 11 bool  prime[100005];
 12 int ans[100005];
 13 int bns[100005];
 14 int dd[100005];
 15 typedef long long LL;
 16 int main(void)
 17 {
 18         int i,j,k;
 19         scanf("%d",&k);
 20         int s;
 21         LL n,m,x;
 22         for(i=2; i<=1000; i++)
 23         {
 24                 if(!prime[i])
 25                 {
 26                         for(j=i; i*j<=100000; j++)
 27                         {
 28                                 prime[i*j]=true;
 29                         }
 30                 }
 31         }
 32         int cnt=0;
 33         for(i=2; i<=100000; i++)
 34         {
 35                 if(!prime[i])
 36                 {
 37                         ans[cnt++]=i;
 38                 }
 39         }
 40         for(s=1; s<=k; s++)
 41         {
 42                 int uu=0;
 43                 memset(dd,0,sizeof(dd));
 44                 scanf("%lld %lld %lld",&n,&m,&x);
 45                 while(x>=1&&uu<cnt)
 46                 {
 47                         if(x%ans[uu]==0)
 48                         {
 49                                 dd[ans[uu]]=1;
 50                                 x/=ans[uu];
 51                         }
 52                         else
 53                         {
 54                                 uu++;
 55                         }
 56                 }
 57                 int qq=0;
 58                 for(i=2; i<=100000; i++)
 59                 {
 60                         if(dd[i])
 61                         {
 62                                 bns[qq++]=i;
 63                         }
 64                 }
 65                 if(x!=1)
 66                         bns[qq++]=x;
 67                 n--;
 68
 69                 LL nn=0;
 70                 LL mm=0;
 71                 for(i=1; i<=(1<<qq)-1; i++)
 72                 {
 73                         int xx=0; LL sum=1;
 74                         int flag=0;
 75                         for(j=0; j<qq; j++)
 76                         {
 77                                 if(i&(1<<j))
 78                                 {
 79                                         xx++;
 80                                         LL cc=gcd(sum,bns[j]);
 81                                         sum=sum/cc*bns[j];
 82                                         if(sum>m)
 83                                         {
 84                                                 flag=1;
 85                                                 break;
 86                                         }
 87                                 }
 88                         }
 89                         if(flag)
 90                                 continue;
 91                         else
 92                         {
 93                                 if(xx%2==0)
 94                                 {
 95                                         nn-=n/sum;
 96                                         mm-=m/sum;
 97                                 }
 98                                 else
 99                                 {
100                                         nn+=n/sum;
101                                         mm+=m/sum;
102                                 }
103                         }
104                 }m-=mm;n-=nn;
105                 printf("Case #%d: ",s);
106                 printf("%lld\n",m-n);
107         }
108         return 0;
109 }
110 long long  gcd(long long  n,long long  m)
111 {
112         if(m==0)
113                 return n;
114         else if(n%m==0)
115                 return m;
116         else return gcd(m,n%m);
117 }