ms是莫比乌斯反演里最水的题。。。
题意：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a，y<=b，并且gcd(x,y)=d。
多组询问， T<=50000,d,a,b<=50000
稍微推下shizi
\(\sum_{i=1}^a\sum_{j=1}^b[\gcd(i,j)=d]\)
\(=\sum_{i=1}^{a/k}\sum_{j=1}^{b/k}[\gcd(i,j)=1]\)
\(=\sum_{i=1}^{a/k}\sum_{j=1}^{b/k}\sum_{d|i,d|j}\mu(d)\)
\(=\sum_{i=1}^{a/k}\sum_{j=1}^{b/k}\sum_{d|i,d|j}\mu(d)\)
\(=\sum_{d=1}^n\mu(d)\lfloor\frac a{kd}\rfloor\lfloor\frac b{kd}\rfloor\)
连枚举倍数都不用。。。直接打个数论分块就行了。。。复杂度\(O(T\sqrt n)\)
#include <cstdio>
#include <functional>
using namespace std;
bool visit[50010];
int prime[50010], mu[50010], tot, fuck = 50000;
int main()
{
mu[1] = 1;
for (int i = 2; i <= fuck; i++)
{
if (visit[i] == false) prime[++tot] = i, mu[i] = -1;
for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
{
visit[i * prime[j]] = true;
if (i % prime[j] == 0) break;
mu[i * prime[j]] = -mu[i];
}
mu[i] += mu[i - 1];
}
int t;
scanf("%d", &t);
while (t --> 0)
{
int n, m, k;
scanf("%d%d%d", &n, &m, &k);
n /= k, m /= k;
int res = 0;
if (n > m) swap(n, m);
for (int i = 1, j; i <= n; i = j + 1)
{
j = min(n / (n / i), m / (m / i));
res += (mu[j] - mu[i - 1]) * (n / i) * (m / i);
}
printf("%d\n", res);
}
return 0;
}
40行一遍AC
于NOIWC2019 Day2晚试机
日推里出现的题，竟然挺水
NOI Linux真TM难用，累死我了