HDU 2588 - GCD (欧拉函数)
题意
找出1~n之间且gcd(x, n) >= M的个数。
思路
假设$X$是$N$的一个约数, 且$X >= M$, 那么设$P = \frac {N}{X}$
那么显然$gcd(X*k, N) = X, \text {1 <= k < P && gcd(k, P) = 1}$
所以$ans = \sum \phi (\frac {N}{k}), \text {k是N的约数且k>=M} $
代码
#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
const int hash_size = 4e5 + 10;
int cases = 0;
typedef pair<int, int> pii;
int ans, M, N;
vector<int> fac;
map<int, int> mp;
int get_phi(int n)
{
int m = sqrt(n+0.5);
int aans = n;
for (int i = 2; i <= m; i++) if (n%i == 0)
{
aans = aans / i * (i-1);
while (n % i == 0) n /= i;
}
if (n > 1) aans = aans / n * (n-1);
return aans
}
void get_fac(int n)
{
int m = (int)sqrt(n+0.5);
for (int i = 1; i <= m; i++)
{
if (n % i) continue;
if (i >= M && i != m)
ans += get_phi(N / i);
if (N / i >= M) ans += get_phi(i);
}
}
int Solve(int n)
{
ans = 0;
get_fac(n);
}
int main()
{
//ROP;
int T;
scanf("%d", &T);
while (T--)
{
int n;
scanf("%d%d", &N, &M);
Solve(N);
printf("%d\n", ans);
}
return 0;
}