Visible Trees
解题思路:
实际上的答案就是1~n与1~m之间互质的数的对数，写出式子就是
(ans=sum^{n}_{i=1}sum^{m}_{j=1}[gcd(i,j)=1])
由莫比乌斯反演引理
(sum_{d|n}mu(d)=epsilon(n)=[n=1])将(epsilon(n))替换为([gcd(i,j)=1])有
(sum_{d|gcd(i,j)}mu(d)=[gcd(i,j)=1])
(ans=sum^{n}_{i=1}sum^{m}_{j=1}[gcd(i,j)=1]=sum^{n}_{i=1}sum^{m}_{j=1}sum_{d|gcd(i,j)}mu(d))
现在枚举(d)
由于(d)同时是(i,j)的因子
(ans=sum^n_{d=1}mu(d)*lfloorfrac{n}{d}
floorlfloorfrac{m}{d}
floor)
后面(mu(d)*lfloorfrac{n}{d}
floorlfloorfrac{m}{d}
floor)能数论分块做,复杂度(O(sqrt{n}))
还是挺套路的
具体实现
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
// clock_t c1 = clock();
// std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define ls ((x) << 1)
#define rs ((x) << 1 | 1)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<ll, ll> PLL;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e6 + 7;
const ll MAXM = 1e5 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
ll mu[MAXN], pri[MAXN], vis[MAXN], tot = 0;
ll sum[MAXN];
void init()
{
mu[1] = 1;
for (int i = 2; i < MAXN; i++)
{
if (!vis[i])
pri[++tot] = i, mu[i] = -1;
for (int j = 1; j <= tot && pri[j] * i < MAXN; j++)
{
vis[i * pri[j]] = 1;
if (i % pri[j] == 0)
mu[i * pri[j]] = 0;
else
mu[i * pri[j]] = -mu[i];
}
}
for (int i = 1; i < MAXN; i++)
sum[i] = sum[i - 1] + mu[i];
}
ll go(int n, int m)
{
ll ans = 0;
int last = 0;
for (int l = 1; l <= n; l = last + 1)
{
last = min((n / (n / l)), (m / (m / l)));
ans += (sum[last] - sum[l - 1]) * (n / l) * (m / l);
}
return ans;
}
int main()
{
init();
int t;
scanf("%d", &t);
while (t--)
{
int n, m;
scanf("%d%d", &n, &m);
if (n > m)
swap(n, m);
printf("%lld
", go(n, m));
}
return 0;
}