51Nod-1227-平均最小公倍数

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ACM模版

描述

题解

默比乌斯反演 + 杜教筛 + 分块 + 欧拉函数！！！这个套路的题 51Nod 上真多……

HOWARLI’s blog 可供详细参考！！！这几道套路题大佬差不多都写了，我就是看看大佬的题解拓展拓展眼界~~~

代码

#include <cstdio>#include <cstdlib>using namespace std;typedef long long ll;const int MAXN = 4e6;const int MOD = 1e9 + 7;const int MAXM = 3e5;const int INV_2 = 5e8 + 4;const int INV_6 = 166666668;int n, m;bool prz[MAXN + 10];int pri[MAXN >> 1];int phi[MAXN + 10];int hx[MAXM][2];int HX(int q){    int tmp = q % MAXM;    while (hx[tmp][0] && hx[tmp][0] != q)    {        tmp = (tmp + 1) % MAXM;    }    return tmp;}ll SM(ll s, ll t){    return (s + t) * (t - s + 1) % MOD * INV_2 % MOD;}ll Gphi(int q){    if (q <= MAXN)    {        return phi[q];    }    int t = HX(q);    if (hx[t][0])    {        return hx[t][1];    }    hx[t][0] = q;    ll ans = 0;    for (int i = 2, nx; i <= q; i = nx + 1)    {        nx = q / (q / i);        ans = (ans + SM(i, nx) * Gphi(q / i) % MOD) % MOD;    }    q %= MOD;    return hx[t][1] = (int)((ll)q * (q + 1) % MOD * (2 * q + 1) % MOD * INV_6 % MOD - ans);}ll Gans(int n){    int ans = n % MOD;    for (int i = 2, nx; i <= n; i = nx + 1)    {        nx = n / (n / i);        ans = (ans + (ll)(n / i) * (Gphi(nx) - Gphi(i - 1)) % MOD * INV_2 % MOD) % MOD;    }    return ans;}void init(){    phi[1] = 1;    for (int i = 2; i <= MAXN; i++)    {        if (!prz[i])        {            pri[++pri[0]] = i;            phi[i] = i - 1;        }        for (int j = 1; j <= pri[0]; j++)        {            int t = pri[j] * i;            if (t > MAXN)            {                break;            }            prz[t] = 1;            phi[t] = phi[i] * pri[j];            if (i % pri[j] == 0)            {                break;            }            phi[t] = phi[i] * (pri[j] - 1);        }    }    for (int i = 2; i <= MAXN; i++)    {        phi[i] = ((ll)phi[i] * ((ll)i) % MOD + phi[i - 1]) % MOD;    }}int main(){    init();    scanf("%d%d", &m, &n);    printf("%lld\n", (Gans(n) - Gans(m - 1) + MOD) % MOD);    return 0;}

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