Codeforces 599D：Spongebob and Squares

D. Spongebob and Squares

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs of n and m, such that there are exactly x distinct
squares in the table consisting of n rows and m columns.
For example, in a 3 × 5 table there are 15squares
with side one, 8 squares with side two and 3 squares
with side three. The total number of distinct squares in a 3 × 5 table is15 + 8 + 3 = 26.

Input

The first line of the input contains a single integer x (1 ≤ x ≤ 1018) —
the number of squares inside the tables Spongebob is interested in.

Output

First print a single integer k — the number of tables with exactly x distinct
squares inside.

Then print k pairs of integers describing the tables. Print the pairs in the order of increasing n,
and in case of equality — in the order of increasing m.

Sample test(s)

input

26

output

6
1 26
2 9
3 5
5 3
9 2
26 1

input

2

output

2
1 2
2 1

input

8

output

4
1 8
2 3
3 2
8 1

Note

In a 1 × 2 table there are 2 1 × 1 squares.
So, 2 distinct squares in total.

In a 2 × 3 table there are 6 1 × 1 squares
and 2 2 × 2 squares. That
is equal to 8 squares in total.

题意是给定一个X，问那些矩形中含有的正方形总数等于X。

这题当时没时间做了，(太弱。。。)后面补的。

官方题解：

第一点：n*m里面的正方形数量就是sum((n-i)*(m-i))，i从1到n-1啊。。。在纸上画几次就明白了。

第二点：从1到n的平方和等于n(n+1)(2n+1)/6。。。

然后就是枚举n，求m。

代码：

#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <map>
using namespace std;
typedef long long ll;

const int maxn = 2000005;
ll x;
ll a[maxn];
ll b[maxn];

int main()
{
    //freopen("i.txt", "r", stdin);
    //freopen("o.txt", "w", stdout);

    int flag;
    ll i, len, num, n, m, temp;
    cin >> x;

    flag = -1;
    num = 0;
    len = 2 * pow((double)x, ((double)1 / (double)3));
    for (i = 1; i <= len+1; i++)
    {
        temp = 6 * x + i*i*i - i;
        n = i*i + i;

        if ((temp % (3 * n) == 0) && (i <= temp / (3 * n)))
        {
            a[num] = i;
            b[num] = temp / (3 * n);

            if (a[num] == b[num])
            {
                flag = num;
            }
            num++;
        }
    }
    if (flag == -1)
    {
        cout << num * 2 << endl;
        for (i = 0; i < num; i++)
        {
            cout << a[i] << " " << b[i] << endl;
        }
        for (i = num-1; i >= 0; i--)
        {
            cout << b[i] << " " << a[i] << endl;
        }
    }
    else
    {
        cout << num * 2 - 1 << endl;
        for (i = 0; i < num; i++)
        {
            cout << a[i] << " " << b[i] << endl;
        }
        for (i = num - 1; i >= 0; i--)
        {
            if (flag == i)
                continue;
            cout << b[i] << " " << a[i] << endl;
        }
    }
    //system("pause");
    return 0;
}