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 <= 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

 

   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.

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#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 1000100using 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;}