hdu 4135 Co-prime(容斥)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2307    Accepted Submission(s): 861

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

 

题意：求A到B之间的数有多少个与n互质。

首先转化为（1---B）与n互质的个数减去（1--- A-1）与n互质的个数

然后就是求一个区间与n互质的个数了，注意如果是求（1---n）与n互质的个数，可以用欧拉函数，但是这里不是到n，所以无法用欧拉函数。

这里用到容斥原理，即将求互质个数转化为求不互质的个数，然后减一下搞定。

求互质个数的步骤：

1、先将n质因数分解

2、容斥原理模板求出不互质个数ans

3、总的个数减掉不互质个数就得到答案

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<set>
 #include<vector>
 using namespace std;
 #define ll long long
 #define N 1000000
 ll A,B,n;
 vector<ll> v;
 ll solve(ll x,ll n)
 {
     v.clear();
     for(ll i=;i*i<=n;i++) //对n进行素数分解
     {
         if(n%i==)
         {
             v.push_back(i);
             while(n%i==)
                n/=i;
         }
     }
     if(n>) v.push_back(n);

     ll ans=;
     for(ll i=;i<( <<v.size() );i++)//用二进制来1,0来表示第几个素因子是否被用到,如m=3，三个因子是2,3,5，则i=3时二进制是011，表示第2、3个因子被用到
     {
         ll sum=;
         ll tmp=;
         for(ll j=;j<v.size();j++)
         {
             if((<<j)&i) //判断第几个因子目前被用到
             {
                 tmp=tmp*v[j];
                 sum++;
             }
         }
         if(sum&) ans+=x/tmp;//容斥原理，奇加偶减
         else ans-=x/tmp;
     }
     return x-ans;
 }
 int main()
 {
     int t;
     int ac=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%I64d%I64d%I64d",&A,&B,&n);
         printf("Case #%d: ",++ac);
         printf("%I64d\n",solve(B,n)-solve(A-,n));
     }
     return ;
 } 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 #define ll long long
 #define N 1000000
 ll A,B,n;
 ll fac[N];
 ll solve(ll x,ll n)
 {
     ll num=;
     for(ll i=;i*i<=n;i++)
     {
         if(n%i==)
         {
             fac[num++]=i;
             while(n%i==)
                n/=i;
         }
     }
     if(n>) fac[num++]=n;

     ll ans=;
     for(ll i=;i<(<<num);i++)
     {
         ll sum=;
         ll tmp=;
         for(ll j=;j<num;j++)
         {
             if((<<j)&i)
             {
                 tmp=tmp*fac[j];
                 sum++;
             }
         }
         if(sum&) ans+=x/tmp;
         else ans-=x/tmp;
     }
     return x-ans;
 }
 int main()
 {
     int t;
     int ac=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%I64d%I64d%I64d",&A,&B,&n);
         printf("Case #%d: ",++ac);
         printf("%I64d\n",solve(B,n)-solve(A-,n));
     }
     return ;
 }