HDU 2588 欧拉函数的应用

最新推荐文章于 2021-06-19 08:38:47 发布

still_foolish

最新推荐文章于 2021-06-19 08:38:47 发布

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分类专栏：数论文章标签：欧拉函数

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本文探讨了如何解决给定整数N和M时，找到满足条件GCD(X,N)≥M的所有整数X的数量的问题。通过使用欧拉函数和枚举因子的方法，有效地解决了这一数学挑战。

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GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1803 Accepted Submission(s): 901

Problem Description
The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.

Input
The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.

Output
For each test case,output the answer on a single line.

Sample Input
3
1 1
10 2
10000 72

Sample Output
1
6
260

Source
ECJTU 2009 Spring Contest

题意：
输入N，M
有多少X满足GCD(X,N)>=M (1<=X<=N)

思路：可以枚举d=GCD(X,N)，可以发现，d一定是N的因子，如果d>=m，那么所有小于等于d且与d互质的数p与N的GCD都是d，所以求出φ(n/d)即可

#include<cstdio>
#include<cmath>
int Euler(int n) {
    if (n == 1)
        return 1;
    int i = 2, m = n, root = (int) sqrt(n);
    while (i <= root) {
        if (m % i == 0) {
            n -= n / i;
            while (m % i == 0)
                m /= i;
            root = (int) sqrt(m);
        }
        i++;
    }
    if (m != 1) {
        n -= n / m;
    }
    return n;
}
int solve(int n, int m) {
    int nn = sqrt(n), ans = 0;
    for (int i = 1; i <= nn; i++) {
        if (n % i)
            continue;
        if (i >= m && i != nn)
            ans += Euler(n / i);
        if (n / i >= m)
            ans += Euler(i);
    }
    return ans;
}
int main() {
    int n, t, m;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        printf("%d\n", solve(n, m));
    }
    return 0;
}