【洛谷】P5348 密碼解鎖

【洛谷】P5348 密碼解鎖

很顯然咱們能夠推導出這個式子

設\(a(m)\)爲\(m\)位置的值
\[ \mu(m) = \sum_{m | d} a(d) \\ a(m) = \sum_{m|d}\mu(\frac{d}{m})\mu(d) \\ a(m) = \sum_{i = 1}^{\lfloor \frac{n}{m} \rfloor} \mu(i)\mu(im) \\ a(m) = \mu(m) \sum_{i = 1}^{\lfloor \frac{n}{m} \rfloor} \mu(i)^{2}[gcd(m,i) == 1] \]
而\(\mu(i)^{2}\)的本質是無平方因子數，這個能夠容斥

容斥的方法是（若沒有其餘限制）
\[ ans = \sum_{i = 1}^{\sqrt{N}} \mu(i)\lfloor \frac{N}{i^{2}}\rfloor \]

那麼這裏的就是
\[ ans = \sum_{i = 1}^{\sqrt{N / M}} [gcd(i,m) == 1]\mu(i)\sum_{j = 1}^{\lfloor \frac{N}{i ^ 2} \rfloor} [gcd(j,m) == 1] \]
前面的互質能夠直接枚舉

後面的互質能夠經過莫比烏斯反演外加預處理M中莫比烏斯值不爲0的數算出來

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int64 N,MAXV;
int M,mu[1000005];
int prime[1000005],tot;
bool nonprime[1000005];
vector<pii > division;
int gcd(int a,int b) {
    return b == 0 ? a : gcd(b,a % b);
}
int Mu(int x) {
    if(x <= 1000000) return mu[x];
    int res = 1;
    for(int i = 2 ; i <= x / i ; ++i) {
    if(x % i == 0) {
        int c = 0;
        while(x % i == 0) {x /= i;++c;}
        if(c >= 2) return 0;
        res = -res;
    }
    }
    if(x != 1) res = -res;
    return res;
}
int64 calc(int64 n) {
    int64 res = 0;
    for(auto t : division) {
    if(n < t.fi) break;
    res += 1LL * t.se * (n / t.fi);
    }
    return res;
}
void Init() {
    mu[1] = 1;
    for(int i = 2 ; i <= 1000000 ; ++i) {
    if(!nonprime[i]) {
        prime[++tot] = i;
        mu[i] = -1;
    }
    for(int j = 1 ; j <= tot ; ++j) {
        if(prime[j] > 1000000 / i) break;
        nonprime[i * prime[j]] = 1;
        if(i % prime[j] == 0) break;
        else mu[i * prime[j]] = -mu[i];
    }
    }
}
void Solve() {
    read(N);read(M);
    if(Mu(M) == 0) {puts("0");return;}
    division.clear();
    for(int i = 1 ; i <= M / i ; ++i) {
    if(M % i == 0) {
        int j = M / i;
        int x = Mu(i),y = Mu(j);
        if(x) division.pb(mp(i,x));
        if(i != j && y) division.pb(mp(j,y));
    }
    }
    sort(division.begin(),division.end());
    int64 T = N / M,res = 0;
    for(int i = 1 ; i <= T / i ; ++i) {
    if(gcd(i,M) == 1) {
        res += Mu(i) * calc(T / (i * i));
    }
    }
    res = res * Mu(M);
    out(res);enter;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    int T;
    read(T);
    for(int i = 1 ; i <= T ; ++i) Solve();
}