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  • P3327 [SDOI2015]约数个数和 莫比乌斯反演

    P3327 [SDOI2015]约数个数和 莫比乌斯反演

    链接

    luogu

    思路

    第一个式子我也不会,luogu有个证明,自己感悟吧。

    [d(ij)=sumlimits_{x|i}sumlimits_{y|j}[gcd(x,y)==1] ]

    [sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}sumlimits_{x|i}sumlimits_{y|j}[gcd(x,y)==1] ]

    [sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}{left lfloor frac{n}{i} ight floor left lfloor frac{m}{j} ight floor left [ gcd(i,j)==1 ight ]} ]

    [f(x)=sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}{left lfloor frac{n}{i} ight floor left lfloor frac{m}{j} ight floor left [ gcd(i,j)==x ight ]} ]

    [g(x)=sumlimits_{x|d} f(d) ]

    [g(x)=sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}{left lfloor frac{n}{i} ight floor left lfloor frac{m}{j} ight floor left [ x|gcd(i,j) ight ]} ]

    [g(x)=sumlimits_{i=1}^{frac{n}{x}}sumlimits_{j=1}^{frac{m}{x}}{left lfloor frac{n}{x*i} ight floor left lfloor frac{m}{x*j} ight floor } ]

    [g(x)=sumlimits_{i=1}^{frac{n}{x}}{left lfloor frac{n}{x*i} ight floor }sumlimits_{j=1}^{frac{m}{x}}{left lfloor frac{m}{x*j} ight floor } ]

    [g(x)=sumlimits_{i=1}^{N}{left lfloor frac{N}{i} ight floor }sumlimits_{j=1}^{M}{left lfloor frac{M}{j} ight floor }(N=n/x,M=m/x) ]

    整除分块预处理,O(1)查询g(x)

    [f(x)=sumlimits_{x|d}mu(frac{d}{n})g(d) ]

    所求$$f(1)=sumlimits_{d=1}^{min(m,n)}mu(d)g(d)$$
    g是可以整除分块的

    其他

    改马蜂,加空格

    代码

    #include <bits/stdc++.h>
    const int N = 5e5+7;
    using namespace std;
    int read() {
    	int x = 0, f = 1; char s = getchar();
    	for (;s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
    	for (;s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    	return x * f;
    }
    int n, m, T;
    int pri[N], vis[N], tot, mu[N], g[N];
    void Euler(int limit) {
    	mu[1] = 1;
    	for (int i = 2; i <= limit; ++i) {
    		if (!vis[i]) {
    			pri[++tot] = i;
    			mu[i] = -1;
    		}
    		for (int j = 1; j <= tot && i * pri[j] <= limit; ++j) {
    			vis[i * pri[j]] = 1;
    			if (i % pri[j] == 0) {
    				mu[i * pri[j]] = 0;
    				break;
    			}
    			mu[i * pri[j]] = -mu[i];
    		}
    	}
    	for (int i = 1; i <= limit; ++i) {
    		for (int l = 1, r; l <= i; l = r + 1) {
    			r = i / (i / l);
    			g[i] += (r - l + 1) * (i / r);
    		}
    		mu[i] += mu[i - 1];
    	}
    }
    void solve() {
    	n = read(), m = read();
    	if (n > m) swap(n, m);
    	long long ans = 0;
    	for (int l = 1, r; l <= n; l = r + 1) {
    		r = min(n / (n / l), m / (m / l));
    		ans += 1LL * (mu[r] - mu[l-1]) * (1LL * g[n/l] * g[m/l]);
    	}
    	printf("%lld
    ", ans);
    }
    int main() {
    	Euler(50000);
    	int T = read();
    	while (T--) solve();
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dsrdsr/p/10804285.html
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