CF1717E_cf1710e-优快云博客

本文链接：https://blog.youkuaiyun.com/CK1513710764/article/details/126676025

该博客主要探讨了如何利用欧拉函数预处理和因数分解来解决数学问题，具体涉及gcd(a,b)的计算，并将其转化为求解互质关系的问题。博主通过C++代码展示了如何高效地计算给定整数n的不同子集之和的互质元素个数，整个过程涉及到了数论中的基本概念和算法，如最小质因子、互质关系和最大公约数的计算。

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思路:

我们假设gcd(a, b) = t,然后求 $\sum_{a + b + c = n} lcm(c, t) = \sum lcm(c, t) \sum_{a + b = n - c}[gcd(a, b) = t]$ ,这里技巧性地构造了 $\sum_{a + b = n - c}[gcd(a, b) = t]$ ,接下来 $gcd(\frac{a}{t},\frac{b}{t}) = 1$ ,然后转化成欧拉函数,接下来找互质即可, $\frac{a}{t} + \frac{b}{t} = \frac{n - c}{t}$ ,显然 $\frac{a}{t}$ 与 $\frac{n - c}{t}$ 互质,因为 $\frac{a}{t}$ 和 $\frac{b}{t}$ 互质,提不出公共的约数.此时 $\sum_{a + b = n - c}[gcd(a, b) = t] = \varphi(\frac{n - c}{t})(\frac{n - c}{t} \neq 1)$ .

我们预处理欧拉函数和用埃式筛法预处理每个数的因数,然后枚举n - c的因数t,然后计算即可.

#include <bits/stdc++.h>
#define int long long                                         
#define IOS ios::sync_with_stdio(false), cin.tie(0) 
#define ll long long 
// #define double long double
#define ull unsigned long long 
#define PII pair<int, int> 
#define PDI pair<double, int> 
#define PDD pair<double, double> 
#define debug(a) cout << #a << " = " << a << endl 
#define point(n) cout << fixed << setprecision(n)
#define all(x) (x).begin(), (x).end() 
#define mem(x, y) memset((x), (y), sizeof(x)) 
#define lbt(x) (x & (-x)) 
#define SZ(x) ((x).size()) 
#define inf 0x3f3f3f3f 
#define INF 0x3f3f3f3f
namespace nqio{const unsigned R = 4e5, W = 4e5; char *a, *b, i[R], o[W], *c = o, *d = o + W, h[40], *p = h, y; bool s; struct q{void r(char &x){x = a == b && (b = (a = i) + fread(i, 1, R, stdin), a == b) ? -1 : *a++;} void f(){fwrite(o, 1, c - o, stdout); c = o;} ~q(){f();}void w(char x){*c = x;if (++c == d) f();} q &operator >>(char &x){do r(x);while (x <= 32); return *this;} q &operator >>(char *x){do r(*x); while (*x <= 32); while (*x > 32) r(*++x); *x = 0; return *this;} template<typename t> q&operator>>(t &x){for (r(y),s = 0; !isdigit(y); r(y)) s |= y == 45;if (s) for (x = 0; isdigit(y); r(y)) x = x * 10 - (y ^ 48); else for (x = 0; isdigit(y); r(y)) x = x * 10 + (y ^ 48); return *this;} q &operator <<(char x){w(x);return *this;}q &operator<< (char *x){while (*x) w(*x++); return *this;}q &operator <<(const char *x){while (*x) w(*x++); return *this;}template<typename t> q &operator<< (t x) {if (!x) w(48); else if (x < 0) for (w(45); x; x /= 10) *p++ = 48 | -(x % 10); else for (; x; x /= 10) *p++ = 48 | x % 10; while (p != h) w(*--p);return *this;}}qio; }using nqio::qio;
using namespace std;
const int N = 2e5 + 10, MOD = 1e9 + 7;
int n, phi[N], primes[N], cnt, vis[N];
vector<int> factor[N];
void get_euler(int n) {
    phi[1] = 1;
    for(int i = 2; i <= n; ++ i) {
        if(!vis[i]) {
            primes[++cnt] = i;
            phi[i] = i - 1;
        }
        for(int j = 1; j <= cnt && i * primes[j] <= n; ++ j) {
            vis[i * primes[j]] = true;
            if(i % primes[j] == 0) {//最小的质因子
                phi[i * primes[j]] = phi[i] * primes[j];
                break;
            }
            phi[i * primes[j]] = phi[i] * (primes[j] - 1);
        }
    }
    phi[1] = 0;
    for (int i = 1; i <= n; ++i)
    	for (int j = 1; j * i <= n; ++j) factor[i * j].emplace_back(j);
}
int lcm(int a, int b) {
	return a / __gcd(a, b) * b;
}
signed main() {
	get_euler(2e5);
	qio >> n;
	int ans = 0;
	for (int c = 1; c < n; ++c)
		for (int t : factor[n - c])
			(ans += lcm(c, t) * phi[(n - c) / t]) %= MOD;
	qio << ans << "\n";
}