HDU 4135 Co-prime [容斥定理]【数论】

题目连接： 传送阵
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Co-prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3902 Accepted Submission(s): 1536

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}.

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题目大意： 就是求a~b区间内与n互质的数的个数

解题思路：与n互质的数的个数也就是gcd(x,n)==1.但是数据量非常大 所以暴力不可解
于是换个思路 就是求gcd(x,n)!=1的数的个数 然后区间总数减一下 就能得到结果
gcd(x,n)!=1就简单了

只要求出[1~a-1][1~b]这两个区间内的与n有约数(非1)的数的个数
想到把n质因子分解 然后容斥定理求解即可

附本题代码
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#include <stdio.h>#include <vector>#include <iostream>using namespace std;#define LL long long int#define pb push_backLL solve (LL n, LL r){    vector<int> p;    for (int  i=2; i*i<=n; ++i)        if (n % i == 0)        {            p.pb (i);            while (n % i == 0)                n /= i;        }    if (n > 1)  p.pb (n);    LL sum = 0;    for (int msk=1; msk<(1<<p.size()); ++msk)    {        LL mult = 1,            bits = 0;        for (int i=0; i<(LL)p.size(); ++i)            if (msk & (1<<i))            {                ++bits;                mult *= p[i];            }        LL cur = r / mult;        if (bits % 2 == 1)            sum += cur;        else            sum -= cur;    }    return r - sum;}int main(){    int _,p=0;    scanf("%d",&_);    while(_--)    {        LL a,b,n;        scanf("%I64d%I64d%I64d",&a,&b,&n);        LL sum = solve(n,b)-solve(n,a-1);        printf("Case #%d: %I64d\n",++p,sum);    }    return 0;}