容斥原理：HDU-4135Co-prime

容斥原理公式：这里就需要用到容斥原理了，公式就是：n/2+n/3+n/5-n/(2*3)-n/(2*5)-n/(3*5)+n/(2*3*5). 求的是多个重合区间的里面的数字个数。

解题心得：

1、一开始很傻很天真，使用遍历然后调用__gcd()来直接怼，但是肯定要超时啊，a，b的范围太大了。

2、求一个数与另一个数是否互质还有一种算法，看这个数是否是另一个数的质因子的倍数（详细算法见：链接：求一个数的质因子），如果是则排除。这样就可以直接使用质因子来筛选就可以了，但是需要的是个数可以直接做除，这样使用的时间就大大的减少了。所以就可以将思路转换求a到b区间的互质数可以使用，0到b区间的互质数减去0-a-1区间的互质数。

3、详细过程：先将一个数的质因子全部放在一个数组之中，看是否是质因子的倍数，这个时候就需要使用到容斥原理，因为不只是简单的将每个互质数组合的倍数减去就是了，有可能有的数重复减去了。

题目：

Co-prime
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4738    Accepted Submission(s): 1894


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
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

代码：

#include<bits/stdc++.h>using namespace std;const int maxn = 1e6;int prim[maxn];//用来存储质因子int ch[maxn];//用来存储质因子的组合int n,T;long long a,b;//得到质因子void prim_num(int N){    T = 0;    for(int i=2;i*i<=N;i++)    {        if(N%i == 0)        {            prim[T++] = i;            while(N%i == 0)                N /= i;        }    }    if(N != 1)        prim[T++] = N;}long long Check(long long num){    memset(ch,0,sizeof(ch));    ch[0] = -1;    int t2 = 1;    for(int i=0;i<T;i++)    {        int now;        now = t2;        //这个循环很重要它是得到的质因子的组合，仔细理解（顺序并不是和公式上面的顺序一样）        for(int j=0;j<now;j++)            ch[t2++] = ch[j]*prim[i]*(-1);    }    long long sum = 0;    for(int j=1;j<t2;j++)        sum = sum + num/ch[j];//虽然看起来都是加，但是有正有负，得到的就是最终的答案    return sum;}int main(){    int t;    scanf("%d",&t);    int z = t;    while(t--)    {        scanf("%lld%lld%d",&a,&b,&n);        prim_num(n);        printf("Case #%d: ",z-t);        long long now1 = b - Check(b);        long long now2 = a-1 - Check(a-1);        printf("%lld\n",now1 - now2);    }}