ZOJ3558 How Many Sets III(公式题)

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How Many Sets III

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given a set S = {1, 2, ..., n}, your job is to count how many set T satisfies the following condition:

• T is a subset of S
• Elements in T can form an arithmetic progression in some order.

Input

There are multiple cases, each contains only one integer n ( 1 ≤ n ≤ 10⁹ ) in one line, process to the end of file.

Output

For each case, output an integer in a single line: the total number of set T that meets the requirmentin the description above, for the answer may be too large, just output it mod 100000007.

Sample Input

2
3

Sample Output

1
4

看到这种输出只和一个数有关的，而且还是整数，想都不想，先暴力求出前几项，然后oeis大法，查到公式后

a(n) = sum { i=1..n-1, j=1..floor((n-1)/i) } (n - i*j)

发现这个公式只是n^2的，于是我们需要优化其中的步骤，首先，对于第二维，我们很容易搞掉，那么对于第一维，我们发现其中有一个(n-1)/i，那么其实有很多是对应的，于是我们只需要枚举1到sqrt(n-1)即可。即对于每一个i，在公差在(n-1)/(i+1) + 1到(n-1)/i这个范围内是可求的，另外注意求一下其相对的情况，看上去比较轻松，然而我这种数学渣还是推了半个多小时才推出来的

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author xyiyy @https://github.com/xyiyy
 */

#include <iostream>
#include <fstream>

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>

using namespace std;
typedef long long ll;

//
// Created by xyiyy on 2015/8/5.
//

#ifndef ICPC_INV_HPP
#define ICPC_INV_HPP
typedef long long ll;

void extgcd(ll a, ll b, ll &d, ll &x, ll &y) {
    if (!b) {
        d = a;
        x = 1;
        y = 0;
    }
    else {
        extgcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
}

ll inv(ll a, ll mod) {
    ll x, y, d;
    extgcd(a, mod, d, x, y);
    return d == 1 ? (x % mod + mod) % mod : -1;
}


#endif //ICPC_INV_HPP

const ll mod = 100000007;

class TaskJ {
public:
    void solve(std::istream &in, std::ostream &out) {
        ll n;
        while (in >> n) {
            ll ans = 0;
            ll m = n - 1;
            ll num = inv(2, mod);
            for (ll i = 1; i * i <= m; i++) {
                ll r = m / i;
                ll l = m / (i + 1) + 1;
                if (l > r)continue;
                ans += (n * i % mod * (r - l + 1) % mod -
                        (1LL + i) * i % mod * num % mod * (l + r) % mod * (r - l + 1) % mod * num % mod) % mod + mod;
                ans %= mod;
                if (i != r)ans += (n * r % mod - i * r % mod * (1LL + r) % mod * num % mod) % mod + mod;
                ans %= mod;
            }
            out << ans << endl;
        }
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    TaskJ solver;
    std::istream &in(std::cin);
    std::ostream &out(std::cout);
    solver.solve(in, out);
    return 0;
}