HDU4135容斥原理

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8481    Accepted Submission(s): 3374

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

 题解：容斥原理

求[a,b]中与k互质的个数，即求[1,b]与k互质的个数-[1,a-1]与k互质的个数。

求区间[1,n]与k互质的个数，我们先求[1,n]与k不是互质的数的个数。

我们先把k的质因数全部求出来，用数组p[]来保存，质因数个数即为num

接下来没有质因数的组合情况，这里用二进制状态压缩来枚举。

举个栗子?求[1,20]中与6不互质的数的个数

结果显而易见sum = 20 / 2 + 20 / 3 - 20 / 6;

我们已经求得p[0]=  2,p[1] = 3;  num = 2;

所以共有1<

    当i = 1，01，选择质因数2， sum += 20/2=10;
    当i = 2，10，选择质因数3， sum += 20/3=6；
    当i = 3，11，选择质因数2和3， sum -= 20/6 =3;

我们发现奇数加，偶数减，所以用一个bit来记录

具体代码如下

#include 
#include 
using namespace std;
typedef long long LL;
int const N = 1e5;
LL a,b,n,cnt,T,p[N];
LL solve(LL r,int n){
    LL sum = 0,num = 0;
    for(int i=2;i*i<=n;i++){
        if(n%i == 0){
            p[num++] = i;
            while(n%i == 0) n /= i;
        }
    }
    if(n>1) p[num++] = n;
    for(LL msk=1;msk<(1<>T;
    while(T--){
        cin>>a>>b>>n;
        printf("Case #%d: ",++cnt);
        cout<