【51nod】1602 矩阵方程的解

这个行向量显然就是莫比乌斯函数啦，好蠢的隐藏方法= =

然后我们尝试二分，二分的话要求一个这个东西

\(H(n) = \sum_{i = 1}^{n} \mu(i) == d\)

当然\(\mu(x)\)由于一些很好的性质，这个东西可以用分类讨论做出来

众所周知，求\(\mu\)不为0的数的方法就是容斥求无平方因子数

\(G(n) = \sum_{i = 1}^{\sqrt{N}} \mu(i) \lfloor \frac{N}{i^{2}} \rfloor\)

这样我们已经可以快速求得\(\mu(i) = 0\)的位置有多少了，就是\(N - G(n)\)

再求一个\(F(n)\)作为莫比乌斯函数的前缀和，也就是1和-1的差值

可以用\(F(n)\)和\(G(n)\)和一点特判快速求出来1和-1的个数，显然\(F(n)\)需要的复杂度 远大于\(G(n)\)，是\(O(n^{2/3})\)的

梦想选手快乐的写了一发，只过了d = 0的点，惨兮兮

然后经过努力的尝试，发现如果\(R - L < 10^{6}\)的时候停止二分，转而改为通过类似埃氏筛的方法暴力求这\(10^{6}\)个位置的mu函数，然后挨个枚举看看哪里是答案

这样的话可以在3s左右通过了

#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 1000005
//#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);
}
int prime[10000005],tot,mu[10000005],M[10000005];
int c[10000005];
int64 val[10000005];
bool nonprime[10000005];
struct hash {
    struct node {
        int64 x,val;int nxt;
    }E[10000005];
    int sumE,head[1000000],mo = 974711;
    void add(int64 x,int64 v) {
        int u = x % mo;
        E[++sumE].x = x;
        E[sumE].val = v;
        E[sumE].nxt = head[u];
        head[u] = sumE;
    }
    int64 Query(int64 x) {
        int u = x % mo;
        for(int i = head[u] ; i ; i = E[i].nxt) {
            if(E[i].x == x) return E[i].val;
        }
        return -1e18;
    }
}hsh;
map<int64,int64> zz;
int64 F(int64 n) {
    if(n <= 10000000) return M[n];
    //if(zz.count(n)) return zz[n];
    int64 t = hsh.Query(n);
    if(t > -1e18) return t;
    int64 res = 1;
    for(int64 i = 2 ; i <= n ; ++i) {
        int64 r = n / (n / i);
        res -= F(n / i) * (r - i + 1);
        i = r;
    }
    //zz[n] =res;
    hsh.add(n,res);
    return res;
}
int64 G(int64 n) {
    int64 ans = 0;
    for(int64 i = 1 ; i <= n / i ; ++i) {
        ans += 1LL * mu[i] * (n / (i * i));
    }
    return ans;
}
int getmu(int64 x) {
    int r = 1;
    for(int i = 1 ; prime[i] <= x / prime[i] ; ++i) {
        if(x % prime[i] == 0) {
            if(x % (prime[i] * prime[i]) == 0) return 0;
            r = -r;
            x /= prime[i];
        }
    }
    if(x != 1) r = -r;
    return r;
}
int64 check(int64 n,int d) {
    int64 g = G(n);
    if(d == 0) return n - g;
    int64 f = F(n);
    int64 t = (g - abs(f)) / 2;
    if(d == -1 && f < 0) t += abs(f);
    if(d == 1 && f > 0) t += f;
    return t;
}
int main(){
    #ifdef ivorysi
    freopen("f1.in","r",stdin);
    #endif
    mu[1] = 1;M[1] = 1;
    for(int i = 2 ; i <= 10000000 ; ++i) {
        if(!nonprime[i]) {
            prime[++tot] = i;
            mu[i] = -1;
        }
        for(int j = 1 ; j <= tot ; ++j) {
            if(prime[j] > 10000000 / i) break;
            nonprime[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
            else mu[i * prime[j]] = -mu[i];
        }
        M[i] = M[i - 1] + mu[i];
    }
    int64 d,k;
    read(d);read(k);
    int64 L = 1,R = 1e10;

    while(R - L > 1e6) {
        int64 mid = (L + R) >> 1;
        int64 x = check(mid,d);
        if(x >= k) R = mid - 2 * (x - k);
        else L = mid + 2 * (k - x);
    }
    for(int i = 0 ; i <= R - L ; ++i) {c[i] = 1;val[i] = i + L;}
    for(int i = 1 ; i <= tot ; ++i) {
        if(1LL * prime[i] * prime[i] > R) break;
        int64 l = ((L - 1)/ prime[i] + 1) * prime[i];
        int64 r = (R / prime[i]) * prime[i];
        for(int64 j = l - L ; j <= r - L ; j += prime[i]) {
            if(!c[j]) continue;
            if(val[j] % (1LL * prime[i] * prime[i]) == 0) c[j] = 0;
            else {c[j] = -c[j];val[j] /= prime[i];}
        }
    }
    int64 x = check(L - 1,d);
    for(int i = 0 ; i <= R - L ; ++i) {
        if(c[i] && val[i] != 1) c[i] = -c[i];
        if(c[i] == d) ++x;
        if(x == k) {out(i + L);enter;break;}
    }
    //printf("%.3lf\n",1.0 * clock() / CLOCKS_PER_SEC);
    return 0;
}