HDU 4135 Co-prime (容斥原理)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4779    Accepted Submission(s): 1914

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

21 10 23 15 5

 

Sample Output

Case #1: 5Case #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

 

Recommend

lcy

 

题意:求A-B之间与N互质的数的数量

思路:先求出N的因子，之后使用容斥原理

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;typedef long long ll;const int MAXN=1e5+5;int prime[MAXN][10];int cnt[MAXN];int factor[MAXN];int q_factor(ll n){    int num = 0;    for(ll i = 2; i*i <= n; ++i)    {        if(n%i==0)        {            while(n%i==0)            {                n /= i;            }            factor[num++] = i;        }    }    if(n != 1)        factor[num++] = n;    return num;}ll solve(int m,ll n){    ll res=0;    for(int i=1;i<(1<<m);i++)    {        int num=0;        for(int j=i;j!=0;j>>=1) num+=j&1;        int lcm=1;        for(int j=0;j<m;j++)        {            if(i>>j&1)            {                lcm=lcm/__gcd(lcm,factor[j])*factor[j];                if(lcm>n) break;            }        }        if(num%2==0) res-=(ll)(n/lcm);        else res+=(ll)(n/lcm);    }    return res;}int main(){    int t;    scanf("%d",&t);    int casi=1;    while(t--)    {        ll a,b,c;        scanf("%lld%lld%lld",&a,&b,&c);        int k=q_factor(c);        ll ans=b-solve(k,b);        ans-=a-1-solve(k,a-1);        printf("Case #%d: %lld\n",casi++,ans);    }    return 0;}