BZOJ3512 DZY Loves Math IV

3512: DZY Loves Math IV

Time Limit: 15 Sec  Memory Limit: 128 MB

Description

给定n,m，求 模10^9+7的值。

Input

仅一行，两个整数n,m。

Output

仅一行答案。

Sample Input

100000 1000000000

Sample Output

857275582
数据规模：
1<=n<=10^5，1<=m<=10^9，本题共4组数据。

---

数论神题系列，成功卡了3天。。。

首先n很小，直接枚举

然后根据\[ \varphi(in) = \varphi(i) \cdot \varphi(\frac{n}{d})\cdot \sum_{e\mid d} \varphi(e) = \varphi(i) \cdot \sum_{e\mid d}\varphi(\frac{n}{e}) \]这个式子

可以推得\[ S(n,m) = \sum_{d\mid n}\varphi(\frac{n}{d})\cdot S(d, \lfloor \frac{m}{d} \rfloor) \]

---

#include<bits/stdc++.h>
using namespace std;
template <class _T> inline void read(_T &_x) {
    int _t; bool flag = false;
    while ((_t = getchar()) != '-' && (_t < '0' || _t > '9')) ;
    if (_t == '-') _t = getchar(), flag = true; _x = _t - '0';
    while ((_t = getchar()) >= '0' && _t <= '9') _x = _x * 10 + _t - '0';
    if (flag) _x = -_x;
}
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
const int M = 7e6 + 10;
typedef long long LL;
inline int add(int a, int b) {
    a += b;
    if (a < 0) a += mod; if (a >= mod) a -= mod;
    return a;
}
inline int mul(int a, int b) {
    return (int)((LL)a * b % mod);
}
int phi[M], prime[M / 10], pcnt;
int Div[N], mx[N];
bool vis[M];
inline void init() {
    phi[1] = mx[1] = Div[1] = 1;
    for (register int i = 2, j, tmp; i < M; ++i) {
        if (!vis[i]) {
            prime[++pcnt] = i;
            phi[i] = i - 1;
            if (i < N) Div[i] = 1, mx[i] = i;
        }
        for (j = 1; j <= pcnt && (LL)prime[j] * i < M; ++j) {
            tmp = prime[j] * i;
            vis[tmp] = true;
            if (tmp < N) mx[tmp] = prime[j];
            if (i % prime[j] == 0) {
                phi[tmp] = phi[i] * prime[j];
                if (tmp < N) Div[tmp] = Div[i] * prime[j];
                break;
            }
            phi[tmp] = phi[i] * (prime[j] - 1);
            if (tmp < N) Div[tmp] = Div[i];
        }
    }
    for (register int i = 2; i < M; ++i) phi[i] = add(phi[i], phi[i - 1]);
}
struct Hash_table {
    int tot, fir[10000];
    struct Edge {
        int v, res, nxt;
    }e[100000];
    Hash_table() {memset(fir, -1, sizeof fir); tot = 0; }
    inline void add(int a, int b, int c) {
        e[++tot] = (Edge){b, c, fir[a]};
        fir[a] = tot;
    }
    inline int find(int x) {
        int w = x >> 20;
        for (int u = fir[w]; ~u; u = e[u].nxt)
            if (e[u].v == x) return e[u].res;
        return -1;
    }
    inline void insert(int v, int res) {
        add(v >> 20, v, res);
    }
}mp;
int Phi(int n) {
    if (n < M) return phi[n];
    int w = mp.find(n); if (~w) return w;
    int res = (int)(((LL)n * (n + 1) >> 1) % mod);
    for (int i = 2, j; i <= n; i = j + 1) {
        j = n / (n / i);
        res = add(res, -mul(j - i + 1, Phi(n / i)));
    }
    mp.insert(n, res); return res;
}
int sum(int n, int m) {
    if (!m) return 0;
    if (m == 1) return phi[n] - phi[n - 1];
    if (n == 1) return Phi(m);
    int &k = mx[n];
    return add(mul(sum(n / k, m), k - 1), sum(n, m / k));
}
int f[100010];
int main() {
    //freopen("3512.in", "r", stdin);
    //freopen("3512.out", "w", stdout);
    init();
    int n, m, ans = 0; read(n), read(m);
    memset(f, -1, sizeof f);
    for (int i = 1; i <= n; ++i) {
        int k = Div[i], t = i / k, ret;
        if (~f[t]) ret = mul(k, f[t]);
        else ret = mul(k, (f[t] = sum(t, m)));
        ans = add(ans, ret);
    }
    cout << ans << endl;
    return 0;
}