HDU 4135 Co-prime (分解质因数 ＋ 容斥)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2314    Accepted Submission(s): 865

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

21 10 23 15 5

 

Sample Output

Case #1: 5Case #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


题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4135

题目大意：求区间[A,B]内由多少数字与N互质

题目分析：先预处理出N的质因子，开始开的1e5，wa死，开到5e5就过了，算出[A,B]内有多少数字不与N互质，拿总数减即可，计算不互质的数的个数先把区间转化为[1，a-1]和[1，b]然后直接容斥搞

#include <cstdio>#include <cstring>#define ll long longint const MAX = 5e5 + 5;int fac[MAX];int cnt;ll a, b, n, num1, num2;void get_factor(ll x){    cnt = 0;    for(int i = 2; i * i < MAX; i++)    {        if(x % i == 0)        {            fac[cnt ++] = i;            while(x % i == 0)                x /= i;        }    }    if(x > 1)        fac[cnt ++] = x;}   void DFS(int idx, ll cur, int sgin, ll s, bool f){    for(int i = idx; i < cnt; i++)    {        ll tmp = (ll) cur * fac[i];        if(f)            num1 += (ll) sgin * (s / tmp);        else            num2 += (ll) sgin * (s / tmp);        DFS(i + 1, tmp, -sgin, s, f);    }}int main(){    int T;    scanf("%d", &T);    for(int ca = 1; ca <= T; ca++)    {        scanf("%I64d %I64d %I64d", &a, &b, &n);        get_factor(n);        num1 = 0;        DFS(0, 1, 1, b, 1);        num2 = 0;        DFS(0, 1, 1, a - 1, 0);        ll ans = (ll)b - a + 1 - (num1 - num2);        printf("Case #%d: %I64d\n", ca, ans);    }}