HDU 4135 Co-prime（容斥定理）

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2445    Accepted Submission(s): 918

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

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：给出三个整数n,m,k，请输出区间[n,m]中与k互质的个数。

   最简单的容斥定理原理，做了这个差不多就理解一点容斥定理的意思了.

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<vector>
#include<queue>
#include<stack>
#include<map>

#define N 1000100

using namespace std;

__int64 n,m,k;
__int64 prime[N],num[N];
int t;

__int64 IEP(__int64 pn){   /// [n,m]区间求与k互质的个数
    __int64 pt = 0;
    __int64 s = 0;
    num[pt++] = -1;
    for(__int64 i=0;i<t;i++){
        __int64 l = pt;
        for(__int64 j=0;j<l;j++){
            num[pt++] = num[j]*prime[i]*(-1);
        }
    }
    for(__int64 i=1;i<pt;i++){
        s += pn/num[i];
    }
    return s;
}

int main(){
    int T;
    int kk = 0;
    scanf("%d",&T);
    while(T--){
        scanf("%I64d%I64d%I64d",&n,&m,&k);
        __int64 pk = sqrt(k);
        t = 0;
        for(int i=2;i<=pk;i++){
            if(k%i == 0){
                prime[t++] = i;
            }
            while(k%i == 0){
                k = k/i;
            }
        }
        if(k!=1){
            prime[t++] = k;
        }
        __int64 sum = m - n + 1 - IEP(m) + IEP(n-1);
        printf("Case #%d: %I64d\n",++kk,sum);
    }
    return 0;
}