bzoj2005 [Noi2010]能量采集

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2005

【题解】

题目要求$sum_{i=1}^nsum_{j=1}^m ((i,j) * 2 + 1)$

考虑容斥，$t_i$表示有公约数$i$的方案数，显然是$lfloor n/i floor lfloor m/i floor$。

$f_i$表示有最大公约数$i$的方案数，那么有$f_i = t_i - sum_{j=2}^{lfloor n/i floor} f_{ij}$

倒过来做显然就行了，复杂度是$O(nlogn)$。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;

int n, m;
ll f[M], ans; 

int main() {
    cin >> n >> m;
    if(n > m) swap(n, m); 
    for (int i=n; i; --i) {
        f[i] = 1ll * (n/i) * (m/i);
        for (int j=i+i; j<=n; j+=i)
            f[i] -= f[j];
        ans += f[i] * (i+i-1);
    }
    cout << ans; 
    return 0;
}