hdu 4135 Co-prime(容斥原理)

Co-prime

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

HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

不懂容斥原理的话，这有容斥原理的上手题：点击打开链接

有两个原理：
1.在[1,m]之间与m互质的数的个数=[1,m]之间的总个数-[1,m]之间与n不互质的数的个数
2.[1,n]之间与m互质的数的数量 = n - （包含一个质因子的数的个数）+（包含2个质因子的数的个数）-（包含3个质因子的数的个数）+（包含4个质因子的数的个数）.....

代码：

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;

#define LL long long int

vector<int>q;
LL b[1000];

void div(int  k)//对k分解质因子
{
for(int i=2;i*i<=k;i++)
{
if(k%i==0)
q.push_back(i);
while(k%i==0)
k/=i;
}
if(k>1)
q.push_back(k);
}
//容斥原理计算[1,n]内有多少个与k互质的数
LL cont(LL n)
{
LL g=0,sum=n;
b[++g]=1;
for(LL i=0;i<q.size();i++)
{
LL t=g;
for(LL j=1;j<=g;j++)
{
b[++t]=-q[i]*b[j];
sum+=n/b[t];
}
g=t;
}
return sum;
}

int main()
{
int t,m=0;
scanf("%d",&t);
while(m<t)
{
q.clear();
LL a,b;
int k;
scanf("%lld%lld%d",&a,&b,&k);
div(k);
printf("Case #%d: %lld\n",cont(b)-cont(a-1));
}
return 0;
}