hdu4135---Co-prime(容斥原理)

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

Recommend
lcy | We have carefully selected several similar problems for you: 1796 1434 3460 1502 4136

先考虑区间[1, x]里和n不互质的数个数
考虑n的每个素因子
可以被某个素因子整除的数个数为(int)x / pi

但是不能这么单纯算,某些数会被多次统计

用容斥来搞
这个数的素因子不多
所以可以状压来搞出所有组合然后 奇加偶减

/*************************************************************************    > File Name: hdu4135.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年05月26日 星期二 20时37分52秒 ************************************************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace std;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;vector <int> fac;LL calc(LL m) {    int n = fac.size();    LL ans = 0;    for (int i = 1; i < (1 << n); ++i) {        int cnt = 0;        int getf = 1;        for (int j = 0; j < n; ++j) {            if (i & (1 << j)) {                getf *= fac[j];                ++cnt;            }        }        if (cnt % 2) {            ans += (m / getf);        }        else {            ans -= (m / getf);        }    }    return m - ans;}int main() {    int t, icase = 1;    scanf("%d", &t);    while (t--) {        LL l, r;        int n;        cin >> l >> r >> n;        fac.clear();        for (int i = 2; i * i <= n; ++i) {            if (n % i == 0) {                fac.push_back(i);                while (n % i == 0) {                    n /= i;                }            }        }        if (n > 1) {            fac.push_back(n);        }        cout << "Case #" << icase++ << ": " << calc(r) - calc(l - 1) << endl;    }    return 0;}