HDU 2588 欧拉函数的应用

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

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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

1.

#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;}

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