hdu 4135 Co-prime（容斥定理入门）

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}.

解题思路：求区间[A,B]与N互质的数的个数，咱们能够从其对立面来考虑：分别求区间[1,A-1]、区间[1,B]中与N不互质的数的个数为num一、num2，那么区间[A,B]与N互质的数的个数就有(B-num2)-(A-1-num1)。怎么求出区间[1,X]与N不互质的数的个数呢？先分解出N的全部素因子，而后用这些素因子来筛选计算出区间[1,X]中与N不互质的数的个数即X/p_i（p_i为素因子，X为区间右端点），由于任何一个不小于2的数都能表示成若干个素数的乘积，这样就获得区间[1,X]中是素因子的倍数的个数，但为了避免重复和不遗漏计数，应采用容斥定理，公式： ，其中选择某几个素因子能够当作是二进制对应bit上的1，若是当前所选个数为奇数，符号为正，不然为负。注意：容斥计数x/p_i中p_i是几个素数的最小公倍数，因为素数之间是互质的，因此能够直接相乘起来做为其最小公倍数。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 int t,cnt,prime[15];LL a,b,n;
 5 LL solve(LL x){//求与n不互质的总个数
 6     int num;LL ans=0,tp;
 7     for(int i=1;i<(1<<cnt);++i){//用二进制来表示每一个质因子是否被使用，即有2^cnt-1种可能，此时cnt较小，题目中1e9最多也就8个素因子，二进制优化
 8         tp=1,num=0;
 9         for(int j=0;j<cnt;++j)
10             if(i&(1<<j))num++,tp*=prime[j];//表示选择哪几个素因子
11         if(num&1)ans+=x/tp;//奇加
12         else ans-=x/tp;//偶减
13     }
14     return x-ans;
15 }
16 int main(){
17     while(~scanf("%d",&t)){
18         for(int i=1;i<=t;++i){
19             scanf("%lld%lld%lld",&a,&b,&n);cnt=0;
20             for(LL j=2;j*j<=n;++j){//求出n内的全部质因子
21                 if(n%j==0){
22                     prime[cnt++]=j;
23                     while(n%j==0)n/=j;
24                 }
25             }
26             if(n>1)prime[cnt++]=n;
27             printf("Case #%d: %lld\n",i,solve(b)-solve(a-1));//区间差
28         }
29     }
30     return 0;
31 }