HDU 4135：Co-prime （容斥原理）

HDU 4135：Co-prime （容斥原理）

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4289    Accepted Submission(s): 1698

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

21 10 23 15 5

Sample Output

Case #1: 5Case #2: 10

思路：通常我们求1-ｎ中与ｎ互质的数的个数都是用欧拉函数！ 但如果ｎ比较大或者是求１～ｍ中与ｎ互质的数的个数等等问题，要想时间效率高的话还是用容斥原理！容斥：先对n分解质因数，分别记录每个质因数， 那么所求区间内与某个质因数不互质的个数就是n / r(i)，假设r(i)是r的某个质因子 假设只有三个质因子， 总的不互质的个数应该为p1+p2+p3-p1*p2-p1*p3-p2*p3+p1*p2*p3, 及容斥原理，可以转向百度百科查看相关内容 pi代表n/r(i),即与某个质因子不互质的数的个数 ，当有更多个质因子的时候， 可以用状态压缩解决，二进制位上是1表示这个质因子被取进去了。 如果有奇数个1，就相加，反之则相减。

AC代码：

#include#include#include#includetypedef __int64 LL;using namespace std;vectorprime;void jprime(LL n) //分解质因子{ for(LL i=2; i*i<=n; i++) if(n%i==0) { prime.push_back(i); //质因子 while(n%i==0) n/=i; } if(n>1)prime.push_back(n);}LL solve(LL a,LL b,LL n) //[a,b]中与n互质的数的个数{ LL sum=0; LL size=prime.size(); for(LL msk=1; msk<(1<

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