You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
|
|
|
|
|
#include <bits/stdc++.h>
|
|
|
|
|
|
using namespace std;
|
|
|
|
|
|
typedef long long LL;
|
|
|
|
|
|
const int N = 1e4 + 5;
|
|
|
|
|
|
|
|
|
|
|
|
// 例题：HDU4135 HDU2841，HDU1695，HDU3501
|
|
|
|
|
|
|
|
|
|
|
|
LL n, m;
|
|
|
|
|
|
// m 比较大的话可以先打出来素数表再求
|
|
|
|
|
|
|
|
|
|
|
|
// 分解质因数
|
|
|
|
|
|
int p[N], cnt;
|
|
|
|
|
|
void divide(int x) {
|
|
|
|
|
|
cnt = 0;
|
|
|
|
|
|
for (int i = 2; i <= x / i; i++) {
|
|
|
|
|
|
if (x % i == 0) {
|
|
|
|
|
|
p[cnt++] = i;
|
|
|
|
|
|
while (x % i == 0) x /= i;
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
if (x > 1) p[cnt++] = x;
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
LL solve() {
|
|
|
|
|
|
LL ans = 0;
|
|
|
|
|
|
for (int i = 1; i < (1 << cnt); i++) { // 二进制枚举质因子所有组合
|
|
|
|
|
|
LL t = 1, cnt = 0;
|
|
|
|
|
|
for (int j = 0; j < cnt; j++) {
|
|
|
|
|
|
if (i >> j & 1) {
|
|
|
|
|
|
t *= p[j];
|
|
|
|
|
|
cnt++;
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
ans += (cnt & 1 ? n / t : -n / t); // 奇加偶减
|
|
|
|
|
|
}
|
|
|
|
|
|
return n - ans;
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
int main() {
|
|
|
|
|
|
// 求1~10中与3互质的数的个数
|
|
|
|
|
|
/*
|
|
|
|
|
|
测试用例：
|
|
|
|
|
|
10 3
|
|
|
|
|
|
|
|
|
|
|
|
人脑模拟：1,2,4,5,7,8,10 ，共7个
|
|
|
|
|
|
程序输出结果：7
|
|
|
|
|
|
结论：一致，代码应该是正确的
|
|
|
|
|
|
*/
|
|
|
|
|
|
cin >> n >> m;
|
|
|
|
|
|
// 对m分解质因数
|
|
|
|
|
|
divide(m);
|
|
|
|
|
|
|
|
|
|
|
|
cout << solve() << endl;
|
|
|
|
|
|
return 0;
|
|
|
|
|
|
}
|
