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  • hdu 4135 Co-prime (容斥定理)

    Co-prime
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 7679    Accepted Submission(s): 3032

    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}.

    C/C++:

     1 #include <map>
     2 #include <queue>
     3 #include <cmath>
     4 #include <vector>
     5 #include <string>
     6 #include <cstdio>
     7 #include <cstring>
     8 #include <climits>
     9 #include <iostream>
    10 #include <algorithm>
    11 #define INF 0x3f3f3f3f
    12 using namespace std;
    13 
    14 long long t, a, b, n;
    15 
    16 vector <long long> v;
    17 
    18 void get_prime()
    19 {
    20     v.clear();
    21     for (long long i = 2; i * i <= n; ++ i)
    22     {
    23         if (n % i == 0) v.push_back(i);
    24         while (n % i == 0)
    25             n /= i;
    26     }
    27     if (n > 1) v.push_back(n);
    28 }
    29 
    30 long long num(long long m)
    31 {
    32     long long sum = 0, top = 1, Q[1000] = {-1};
    33     for (long long i = 0; i < v.size(); ++ i)
    34     {
    35         long long temp = top;
    36         for (long long j = 0; j < temp; ++ j)
    37             Q[top ++] = -1 * v[i] * Q[j];
    38     }
    39     for (long long i = 1; i < top; ++ i)
    40         sum += (m / Q[i]);
    41     return sum;
    42 }
    43 
    44 int main()
    45 {
    46     scanf("%lld", &t);
    47     for (int i = 1; i <= t; ++ i)
    48     {
    49         scanf("%lld%lld%lld", &a, &b, &n);
    50         get_prime();
    51         printf("Case #%d: %lld
    ", i, b - num(b) - (a - 1 - num(a - 1)));
    52     }
    53 }
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  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9553566.html
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