HDU 4135 Co-prime(组合+容斥）

时间：2015-01-06 23:08:06 收藏：0 阅读：697

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}.

题意：求出[l,r]中n的互素数个数。

容斥+组合：1:我们求出n的所有素因数，然后求出[l,r]之间不互素的数个数s,n-s即答案。

2:n的m个素因数，考虑组合1~(1<<m)-1,容斥可得s.即不互素的数目。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int mod = 1000000007;
LL a,b,n;
int t;
LL solve(LL r,LL n)
{
    vector<int>v;
    for(int i=2;i*i<=n;i++)
    {
        if(n&&n%i==0)
        {
            v.push_back(i);
            while(n&&n%i==0)
                n/=i;
        }
    }
    if(n>1)  v.push_back(n);
    LL sum=0;
    for(int t=1;t<(1<<v.size());t++)
    {
        LL mul=1,bits=0;
        for(int i=0;i<(int)v.size();i++)
        {
            if(t&(1<<i))
            {
                ++bits;
                mul*=v[i];
            }
        }
        if(bits&1)  sum+=r/mul;
        else   sum-=r/mul;
    }
    return r-sum;
}
int main()
{
    int cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d%I64d",&a,&b,&n);
        printf("Case #%d: %I64d\n",cas++,solve(b,n)-solve(a-1,n));
    }
    return 0;
}

原文：http://blog.csdn.net/u013582254/article/details/42472307