[Noi2010] D1T1 能量采集 (欧拉函数)

由结论可知所求数对于i, j来说为gcd(i, j)
推一下公式

ans = n * m + 2 * sigma(i = 1 i <= n) sigma(j = 1 j <= m) gcd(i, j)

由 n = sigma(d | n) phi(d) 得
（略去n * m 及 2） ans = sigma(i = 1 i <= n) sigma(j = 1 j <= m) sigma(d | gcd(i, j)) phi(d)
交换求和顺序
sigma(d = 1 d <= min(n, m)) phi(d) sigma(i = 1 i <= n) [d | i] sigma(j = 1 j <= m) [d | j]
[1, n] 中整除d的个数为n / d 向下取整
式子化为
ans = sigma(d = 1 d <= min(n, m)) phi(d) (n / d)(m / d)
既可以O(n) 也可以O(n ^ 0.5) (分块)

#include <cstdio>#include <cstring>#include <iostream>#include <cmath>#include <algorithm>using namespace std;const int N = 1000005;int n, m, phi[N], pri[N], tot;bool isp[N];void Phi(){    phi[1] = 1;    for(int i = 2; i <= n; ++i)    {        if(!isp[i]) phi[i] = i - 1, pri[++tot] = i;        for(int j = 1; j <= tot; ++j)        {            if(i * pri[j] > n) break;            isp[i * pri[j]] = true;            if(i % pri[j] == 0)            {                phi[i * pri[j]] = pri[j] * phi[i];                break;            }            else phi[i * pri[j]] = phi[i] * (pri[j] - 1);        }    }}int main(){    cin >> n >> m;    if(n < m) swap(n, m);    Phi();    long long ans = 0;    for(int i = 1; i <= min(n, m); ++i)        ans +=  1LL * phi[i] * (n / i) * (m / i);    cout << ans * 2LL - 1LL * n * m << endl;    return 0;}