「LOJ6482」LJJ爱数数
解题思路 :
打表发现两个数 (a, b) 合法的充要条件是(我不管，我就是打表过的)：
[a + b = ext{gcd}(a, b)^2
]
设 (g = ext{gcd(a, b)}) ，那么相当于是要求：
[sum_{g=1}^{sqrt{2n}}sum_{i}[ ext{gcd}(g^2-ig, ig)=g]
]
化简一波：
[sum_{g=1}^{sqrt{2n}}sum_{i}[ ext{gcd}(g-i, i)=1]
]
根据辗转相除：
[sum_{g=1}^{sqrt{2n}}sum_{i}[ ext{gcd}(g, i)=1]
]
考虑 (i) 的上界和下界
[1 leq ig leq n \
1 leq g^2 -ig leq n
]
解一下这两个不等式：
[ ext{max}_i =min(lfloorfrac{n}{g}
floor,g - 1) \
ext{min}_i =max(g-lfloorfrac{n}{g}
floor,1)
]
原来的式子相当于求：
[sum_{g=1}^{sqrt{2n}}sum_{i= ext{min}}^{max}[ ext{gcd}(g, i)=1]
]
设 (f(n)) 表示 ([1, n]) 之间与 (g) 互质的数的个数，反演一波可以得到：
[f(n)= sum_{d|g} lfloor frac{n}{d}
floor mu(d)
]
再化简一波式子：
[sum_{g=1}^{sqrt{2n}}f(max) -f(min-1)
]
总复杂度 (O(sqrt{n}logn)) 。
code
/*program by mangoyang*/
#pragma GCC optimize("Ofast", "inline")
#include<bits/stdc++.h>
#define inf ((int)(1e9))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int f = 0, ch = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 1500005;
int prime[N], d[30000005], tmp[N], mu[N], len[N], b[N], tot;
ll n, ans, m;
inline int solve(int x, int n){
int ans = 0;
for(register int i = len[x-1] + 1; i <= len[x]; i++) ans += mu[d[i]] * (n / d[i]);
return ans;
}
int main(){
read(n), mu[1] = 1;
for(int i = 2; i < N; i++){
if(!b[i]) prime[++tot] = i, mu[i] = -1;
for(int j = 1; j <= tot && i * prime[j] < N; j++){
b[i*prime[j]] = 1;
if(i % prime[j] == 0){ mu[i*prime[j]] = 0; break; }
mu[i*prime[j]] = -mu[i];
}
}
m = (int) sqrt(2ll * n);
for(int i = 1; i <= m; i++) if(mu[i])
for(int j = i; j <= m; j += i) len[j]++;
for(int i = 1; i <= m; i++) len[i] += len[i-1];
for(int i = 1; i <= m; i++) if(mu[i])
for(int j = i; j <= m; j += i) d[(++tmp[j])+len[j-1]] = i;
for(int g = 1; g <= m; g++)
ans += solve(g, Min(n / g, g - 1)) - solve(g, Max(1, g - n / g) - 1);
cout << ans << endl;
return 0;
}