51Nod-1220-约数之和

ACM模版

描述

题解

这个题好难，不会写，给大家提供两个好的博客，一个是我 佐学姐 的博客，讲的是杜教筛的一些总结，《特殊函数的前缀和》，这个写的的确详细，我很喜欢这篇博客，还有一个就是 skywalkert 大佬的博客，讲得这个题的详解，我看了看，半懂半不懂的，真难受，数学底子差做这种题根本就是找不自在……

代码

#include <map>#include <cmath>#include <cstdio>using namespace std;typedef long long ll;const int MAXN = 1e6 + 10;const int MOD = 1e9 + 7;const int INV_2 = 5e8 + 4;int n, sqn, tot, ans;int prime[MAXN];int f[MAXN];bool vis[MAXN];map<ll, int> _hash;inline void inc(int &x, int y){    x += y;    if (x >= MOD)    {        x -= MOD;    }}inline void dec(int &x, int y){    x -= y;    if (x < 0)    {        x += MOD;    }}inline int num1(int x){    return x * (x + 1LL) % MOD * INV_2 % MOD;}inline int num1(int L, int R){    int ret = num1(R);    dec(ret, num1(L - 1));    return ret;}int calc_imu(int x){    if (x <= sqn)    {        return f[x];    }    if (_hash.count(x))    {        return _hash[x];    }    int ret = 1;    for (int i = 2, j; i <= x; i = j + 1)    {        j = x / (x / i);        dec(ret, (ll)num1(i, j) * calc_imu(x / i) % MOD);    }    return _hash[x] = ret;}inline int calc_imu(int L, int R){    int ret = calc_imu(R);    dec(ret, calc_imu(L - 1));    return ret;}int calc_g(int n){    int ret = 0;    for (int i = 1, j; i <= n; i = j + 1)    {        j = n / (n / i);        inc(ret, (j - i + 1LL) * num1(n / i) % MOD);    }    return ret;}int calc_h(int n){    int ret = 0;    for (int i = 1, j; i <= n; i = j + 1)    {        j = n / (n / i);        inc(ret, (ll)(n / i) * num1(i, j) % MOD);    }    return ret;}int main(){    scanf("%d", &n);    f[1] = 1;    sqn = (int)ceil(pow(n, 2.0 / 3));    for (int i = 2; i <= sqn; ++i)    {        if (!vis[i])        {            prime[tot++] = i;            dec(f[i], 1);        }        for (int j = 0, k = sqn / i, o; j < tot && prime[j] <= k; ++j)        {            vis[o = i * prime[j]] = 1;            if (i % prime[j] == 0)            {                f[o] = 0;                break;            }            else            {                dec(f[o], f[i]);            }        }        f[i] = (f[i - 1] + (ll)i * f[i]) % MOD;    }    for (int i = 1, j; i <= n; i = j + 1)    {        j = n / (n / i);        inc(ans, (ll)calc_imu(i, j) * calc_g(n / i) % MOD * calc_g(n / i) % MOD);    }    printf("%d\n", ans);    return 0;}