[数论][容斥原理]Co-prime

Posted 2020-11-03 lz

 

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}.

 

思路：要求区间[a,b]中与n互质的数的个数，可以转化为求[1,b]中与n互质的数的个数和[1,a]中与n互质的数的个数，最后相减即可。

要求[1,b]中与n互质的数的个数，又可以转化为求[1,b]中与n不互质的数的个数，总数减去它便是互质的个数。

要求[1,b]中与n不互质的数的个数，先要清楚怎样的数与n不互质：它必定是n的某个质因子的倍数。

所以先将n分解质因子得到n的若干质因子{fac[1],fac[2],..}，再求出[1,b]中有几个数fac[1]的倍数，几个数fac[2]的倍数，...，再根据容斥原理得到[1,b]中有几个数是{fac[1],fac[2],..}中某数的倍数。

容斥原理：假设n有三个质因子{fac[1],fac[2],fac[3]},则[1,b]中为{fac[1],fac[2],fac[3]}中某数的倍数的数有f(b)=n/fac[1]+n/fac[2]+n/fac[3]-n/(fac[1]*fac[2])-n/(fac[1]*fac[3])-n/(fac[2]*fac[3])+n/(fac[1]*fac[2]*fac[3])个（奇加偶减）

 

AC代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;

ll factor[1000010];
ll que[1000010];
ll cnt;

void get_factor(ll n){//分解质因数(模拟短除法)
  memset(factor,0,sizeof(factor));
  cnt=0;
  for(ll i=2;i*i<=n;i++){
    if(n%i==0){
        factor[++cnt]=i;
        while(n%i==0) n/=i;
    }
  }
  if(n>1) factor[++cnt]=n;
}

ll fun(ll n){//利用数组实现公式的计算
  memset(que,0,sizeof(que));
  ll k=0;
  for(ll i=1;i<=cnt;i++){
    que[++k]=factor[i];
    ll tmp=k;
    for(int j=1;j<=tmp-1;j++) que[++k]=que[tmp]*que[j]*(-1);
  }
  ll ret=0;
  for(ll i=1;i<=k;i++) ret+=n/que[i];
  return ret;
}

int main()
{
    ll t;
    scanf("%lld",&t);
    ll a,b,n;
    for(ll i=1;i<=t;i++){
        scanf("%lld%lld%lld",&a,&b,&n);
        get_factor(n);
        printf("Case #%lld: %lld\n",i,b-fun(b)-(a-1-fun(a-1)));
    }
    return 0;
}