BZOJ3512 DZY Loves Math IV

解：这又是什么神仙毒瘤题......

我直接把后面那个phi用phi * I = id反演一波，得到个式子，然后推不动了......

实际上第一步我就大错特错了。考虑到n很小，我们有

然后计算S，我们根据欧拉函数的性质有：

于是只考虑n sqr free的情况。

到这里有两种解法，一种是暴力递归。

考虑n的一个因子p，我们先提取出前面那一项，但是这还不够。因为当p|i的时候应该提出来p = phi[p] + 1。

于是我们在后面补上。令i = pt，就有了递归式。

#include <cstdio>
#include <map>

typedef long long LL;
const int N = 1000010, T = 1000008;
const LL MO = 1e9 + 7, INF = 0x7f7f7f7f7f7f7f7fll;

int phi[N], p[N], top, last[N];
LL Phi[N], inv2;
bool vis[N];

namespace Hash {
    struct Node {
        LL val, ans;
        int nex;
        Node(LL x = 0, LL y = 0, int z = 0) {
            val = x;
            ans = y;
            nex = z;
        }
    }node[20000010]; int top;
    const int mod = 20000000;
    int e[mod];
    inline void insert(LL p, LL v) {
        int x = p % mod;
        node[++top] = Node(p, v, e[x]);
        e[x] = top;
        return;
    }
    inline LL find(LL p) {
        int x = p % mod;
        for(int i = e[x]; i; i = node[i].nex) {
            if(node[i].val == p) return node[i].ans;
        }
        return -INF;
    }
}

inline void getp(int n) {
    phi[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) {
            p[++top] = i;
            phi[i] = i - 1;
            last[i] = i;
        }
        for(int j = 1; j <= top && i * p[j] <= n; j++) {
            vis[i * p[j]] = 1;
            last[i * p[j]] = p[j];
            if(i % p[j] == 0) {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            phi[i * p[j]] = phi[i] * (p[j] - 1);
        }
    }
    for(int i = 1; i <= n; i++) {
        Phi[i] = (Phi[i - 1] + phi[i]) % MO;
    }
    return;
}

LL getPhi(LL x) {
    if(x <= 0) return 0;
    if(x <= T) return Phi[x];
    LL temp = Hash::find(x);
    if(temp != -INF) return temp;
    //printf("getPhi %lld T = %d
", x, T);
    LL ans = (x % MO) * ((x + 1) % MO) % MO * inv2 % MO;
    for(LL i = 2, j; i <= x; i = j + 1) {
        j = x / (x / i);
        ans -= ((j - i + 1) % MO) * getPhi(x / i) % MO;
        ans = (ans % MO + MO) % MO;
    }
    Hash::insert(x, ans);
    return ans;
}

LL S(LL n, LL m) {
    //printf("S : %lld %lld 
", n, m);
    if(n == 1) return getPhi(m);
    if(m == 0) return 0;
    if(m == 1) return phi[n];
    LL p = last[n];
    LL ans = (phi[p] * S(n / p, m) % MO + S(n, m / p)) % MO;
    return ans;
}

int main() {
    inv2 = (MO + 1) / 2;
    getp(T);
    LL nn, m;
    scanf("%lld%lld", &nn, &m);
    LL ans = 0;
    for(int i = 1; i <= nn; i++) {
        LL p = 1, q = 1, n = i;
        while(n > 1) {
            LL temp = last[n];
            p *= temp;
            n /= temp;
            while(n % temp == 0) {
                n /= temp;
                q *= temp;
            }
        }
        ans = (ans + q * S(p, m) % MO) % MO;
    }
    printf("%lld
", ans);
    return 0;
}

还有一种继续推：

其中d = gcd(i,n)

有个关键点是n是sqr free...否则一直想不明白。

接下来把这个式子带入S的定义式中。

这样就可以递归了。

注意......这个S也可以记忆化的，否则死活过不去。用map。

#include <cstdio>
#include <map>
 
typedef long long LL;
const int N = 1000010, T = 1000008;
const LL MO = 1e9 + 7, INF = 0x7f7f7f7f7f7f7f7fll;
 
int p[N], last[N], top, phi[N];
LL Phi[N], inv2;
bool vis[N];
 
std::map<LL, LL> mp[100010], Hash;
 
inline void getp(int n) {
    phi[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) {
            p[++top] = i;
            phi[i] = i - 1;
            last[i] = i;
        }
        for(int j = 1; j <= top && i * p[j] <= n; j++) {
            vis[i * p[j]] = 1;
            last[i * p[j]] = p[j];
            if(i % p[j] == 0) {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            phi[i * p[j]] = phi[i] * (p[j] - 1);
        }
    }
    for(int i = 1; i <= n; i++) {
        Phi[i] = (Phi[i - 1] + phi[i]) % MO;
    }
    return;
}
 
LL getPhi(LL x) {
    if(x <= 0) return 0;
    if(x <= T) return Phi[x];
    if(Hash.count(x)) return Hash[x];
    //printf("Phi %lld 
", x);
    LL ans = (x % MO) * ((x + 1) % MO) % MO * inv2 % MO;
    for(LL i = 2, j; i <= x; i = j + 1) {
        j = x / (x / i);
        ans -= ((j - i + 1) % MO) * getPhi(x / i) % MO;
        ans = (ans % MO + MO) % MO;
    }
    return Hash[x] = ans;
}
 
LL S(int n, LL m) {
    //printf("S %d %lld 
", n, m);
    if(n == 1) return getPhi(m);
    if(m == 1) return phi[n];
    if(m == 0) return 0;
    if(mp[n][m]) return mp[n][m];
    LL ans = 0;
    for(int i = 1; i * i <= n; i++) {
        if(n % i) continue;
        ans += phi[n / i] * S(i, m / i) % MO;
        if(i * i < n) {
            ans += phi[i] * S(n / i, m / (n / i)) % MO;
        }
        ans = (ans % MO + MO) % MO;
    }
    return mp[n][m] = ans;
}
 
int main() {
    inv2 = (MO + 1) / 2;
    int nn; LL m;
    getp(T);
    scanf("%d%lld", &nn, &m);
    LL ans = 0;
    for(int i = 1; i <= nn; i++) {
        int n = i, p = 1, q = 1;
        //printf("i = %d 
", i);
        while(n > 1) {
            int temp = last[n];
            n /= temp;
            p *= temp;
            while(n % temp == 0) {
                n /= temp;
                q *= temp;
            }
        }
        ans += q * S(p, m) % MO;
        ans = (ans % MO + MO) % MO;
    }
    printf("%lld
", ans);
    return 0;
}

这个时间到底是怎么算的啊......玄学。