HDU4135——Co-prime(数论，容斥原理)

题目：

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3490    Accepted Submission(s): 1379

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 <= 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}. 
   
    
  

 

Source

The Third Lebanese Collegiate Programming Contest

 

题意：求区间[a,b]上与p互质的数字的个数，其中(1 <= a <=b <= 1015) and (1 <=p <= 109)

思路：首先直接求a，b上和p互质的数并不方便。所以考虑分别求[1,a-1],和[1,b]上和p互质的数的个数。它们的差就是答案。

那么如何去求[1,n]上与p互质的数的个数？

这时候就需要用到容斥原理了。以p有三个不同的的质因数为例。设为a，b，c。

那么[1,n]上不和p互质的数的个数为，m=n/a+n/b+n/c-n/(a*b)-n/(a*c)-n/(b*c)+n/(a*b*c)           (奇加偶减)

互质的数的个数就是n-m

计算的时候可以用dfs，队列数组，还有状态压缩。

#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <iostream>

using namespace std;

#define MAXN 55
#define INF 1e9+7
#define MODE 1000000
typedef long long ll;

int t;
ll a,b,n;
vector <int>ans;

ll solve(ll n)
{
    //que用来保存每一项分母，分母是偶数个相乘为负，奇数个为正
    ll que[10000];
    int t=0;
    que[t++]=-1;
    for(int i=0;i<ans.size();i++)
    {
        int k=t;
        for(int j=0;j<k;j++)
        {
            que[t++]=que[j]*ans[i]*(-1);
        }
    }
    ll sum=0;
    for(int i=1;i<t;i++)
        sum=sum+n/que[i];
    return n-sum;
}

int main()
{
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        scanf("%I64d%I64d%I64d",&a,&b,&n);
        ans.clear();
        int temp=n;
        for(int i=2;i*i<=temp;i++)
        {
            if(temp%i==0){
                ans.push_back(i);
                while(temp%i==0)
                    temp/=i;
            }
        }
        if(temp>1)
            ans.push_back(temp);
        ll s1=solve(a-1);
        ll s2=solve(b);
        printf("Case #%d: %I64d\n",cas,s2-s1);
    }
}