hdu4135Co-prime——容斥定理

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}.

---

给你$A,B,N\;$让你找出区间$[A,B]\;$内与$N$互质数的个数
假设$\;g(n)\;$表示从$[1,n]$与n互质数的个数，
那么

$$ans=g(B)-g(A-1)$$
但是我们找 $g(n)$比较麻烦，因为找的都是与n互质的数，我们可以反面思考，1到n内与n互质数的个数就等于n减去与n不互质数的个数

假设$\;f(n)\;$表示从$[1,n]$与n不互质数的个数
那么

$$g(n)=n-f(n)$$
$f(n)$可以根据容斥定理求出来，这样一来，问题就简单化了

---

code:

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;

ll a[1000000+9],factor[1000],sz;

void p(ll n)
{
    sz=0;
    memset(factor,0,sizeof(factor));
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            factor[sz++]=i;
            while(n%i==0)
                n/=i;
        }
    }
    if(n!=1)    factor[sz++]=n;
}
ll f(ll n)//找 1 ... n 的与 n 不互质的数的个数 
{
    ll ans=0;
    for(int i=1;i<(1<<sz);i++)
    {
        int cnt=0;
        ll m=1; 
        for(int j=0;j<sz;j++)
            if((i>>j)&1)
            {
                cnt++;
                m*=factor[j];
            }
        if(cnt&1)   ans += n/m;
        else        ans -= n/m; 
    }
    return ans;
}

int main()
{
    int t;
    ll a,b,n;
    int kase=0; 
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld %lld %lld",&a,&b,&n);
        p(n);
//      for(int i=0;i<sz;i++)
//          printf("%lld ",factor[i]);
//      printf("\n"); 
        ll ans=b-f(b)-(a-1-f(a-1));
        printf("Case #%d: %lld\n",++kase,ans);
    }
    return 0;
}