51nod 1594 Gcd and Phi(莫比乌斯反演)

题目链接

传送门

思路

如果这题是这样的：

\[F(n)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}\phi(gcd(i,j)) \]

那么我们可能会想到下面方法进行反演：

\[\begin{aligned} F(n)=&\sum\limits_{k=1}^{n}\phi(k)\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}[gcd(i,j)=k]&\\ =&\sum\limits_{k=1}^{n}\phi(k)\sum\limits_{i=1}^{\frac{n}{k}}\sum\limits_{j=1}^{\frac{n}{k}}[gcd(i,j)=1]&\\ \end{aligned} \]

令\(f(n)\)为\(gcd(i,j)=n\)的有序对的对数，\(g(n)\)为\(gcd(i,j)=n\text{和}n\)的倍数的有序对的对数，则

\[\begin{aligned} &g(n)=\sum\limits_{n|d}f(d)&\\ \rightarrow&f(n)=\sum\limits_{n|d}\mu(\frac{d}{n})g(d)& \end{aligned} \]

然后将\(f(1)=\sum\limits_{i=1}^{n}\mu(i)g(i)\)代入\(F(n)\)得到\(F(n)=\sum\limits_{k=1}^{n}\phi(k)\sum\limits_{i=1}^{\frac{n}{k}}\mu(i)g(i)\)，然后先预处理\(\sum\limits_{i=1}^{\frac{n}{k}}\mu(i)g(i)\)，然后暴力枚举\(k\)或者数论分块求解答案就可以了，这里有几道类似的题目，有兴趣的可以做一下。

但是这题却不是我们所希望的那种形式，那么该怎么办呢？我们发现\(n\)小于等于\(2e6\)，那么我们可以记录一下每个数的\(\phi\)是多少，然后记录一下每个\(\phi\)出现的次数后就可以转换成上述题意了(这个思想昨天的\(CCPC\)网络赛1010也用到了，不过这题由于后面不会处理就没补了23333)，假设\(c_i\)表示\(\phi=i\)的个数，那么求解的式子就变成了

\[F(n)=\sum\limits_{k=1}^{n}\phi(k)\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}c_ic_j[gcd(i,j)=k] \]

令\(f(n)\)为\(\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}c_ic_j[gcd(i,j)=n]\)，\(g(n)\)为\(为和的倍数\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}c_ic_j[gcd(i,j)\text{为}n\text{和}n\text{的倍数}]\)，则

\[\begin{aligned} g(n)=&\sum\limits_{n|i}\sum\limits_{n|j}c_ic_j&\\ =&\sum\limits_{i=1}^{n}c_i\sum\limits_{j=1}^{n}c_J&\\ =&(\sum\limits_{i=1}^{n}c_i)^2&\\ =&\sum\limits_{n|d}f(d)& \end{aligned} \]

那么

\[f(n)=\sum\limits_{n|d}\mu(\frac{d}{n})g(d) \]

最后答案为

\[F(n)=\sum\limits_{k=1}^{n}\phi(k)f(k) \]

然后暴力枚举\(k\)求解即可。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson (rt<<1),L,mid
#define rson (rt<<1|1),mid + 1,R
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 2000000 + 2;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

bool v[maxn];
int t, n, cnt;
int F[maxn], c[maxn], phi[maxn], p[maxn], mu[maxn];

void init() {
    phi[1] = mu[1] = 1;
    for(int i = 2; i < maxn; ++i) {
        if(!v[i]) {
            p[cnt++] = i;
            phi[i] = i - 1;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && i * p[j] < maxn; ++j) {
            v[i*p[j]] = 1;
            phi[i*p[j]] = phi[i] * (p[j] - 1);
            mu[i*p[j]] = -mu[i];
            if(i % p[j] == 0) {
                mu[i*p[j]] = 0;
                phi[i*p[j]] = phi[i] * p[j];
                break;
            }
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    FIN;
    time_t s = clock();
#endif
    init();
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) {
            ++c[phi[i]];
        }
        F[1] = c[1];
        for(int i = 2; i <= n; ++i) {
            F[1] += c[i];
            for(int j = 1; j <= n / i; ++j) {
                F[i] += c[i*j];
            }
        }
        LL sum = 1LL * F[1] * F[1], ans = 0;
        for(int i = 2; i <= n; ++i) {
            sum += 1LL* F[i] * F[i] * mu[i];
            LL tmp = 0;
            for(int j = 1; j <= n / i; ++j) {
                tmp += 1LL * F[i*j] * F[i*j] * mu[j];
            }
            ans += tmp * phi[i];
        }
        ans += sum;
        printf("%lld\n", ans);
        for(int i = 1; i <= n; ++i) F[i] = c[i] = 0;
    }
#ifndef ONLINE_JUDGE
    printf("It costs %.3fs\n", 1.0 * (clock() - s) / CLOCKS_PER_SEC);
#endif
    return 0;
}