标签：int div bzoj2705 mat cst main for include span
传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2705
令s(k)为gcd(i, n) = k的i的个数，则ans = k * s(k).
若gcd(i, n) = k，则gcd(i / k, n / k) = 1，所以 s(k) = phi(n / k)。 （如果这里没搞懂，随便举个例子就懂了）
剩下的就是O(√n)枚举n的因数就好了~
#include <cstdio>
#include <cmath>
long long n, ans;
int lmt;
int phi(int x) {
int rt = x;
for (int i = 2; i * i <= x; ++i) {
if (x % i == 0) {
rt = rt / i * (i - 1);
x /= i;
while (x % i == 0) {
x /= i;
}
}
}
if (x > 1) {
rt = rt / x * (x - 1);
}
return rt;
}
int main(void) {
scanf("%lld", &n);
lmt = (int)sqrt((float)n + 0.5f);
for (int i = 1; i <= lmt; ++i) {
if (n % i == 0) {
ans += (long long)i * (long long)phi(n / i);
if (n / i != i) {
ans += (long long)(n / i) * (long long)phi(i);
}
}
}
printf("%lld\n", ans);
return 0;
}
标签：int div bzoj2705 mat cst main for include span
原文地址：http://www.cnblogs.com/ciao-sora/p/6368013.html
