【27.40%】【codeforces 599D】Spongebob and Squares

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 15 squares 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 is 15 + 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.

Examples
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.

【题目链接】:http://codeforces.com/problemset/problem/599/D

【题解】


其中(n+m)∑i = (n+m)(0+n-1)*n/2;
而∑i^2 = n*(n+1)*(2n+1)/6
所以x=n^2*m-(m+n)(0+n-1)*n/2+n(n+1)*(2n+1)/6
整理一下得
m= (6*x+n^3-n)/(3*n^2+n);
这样我们只要枚举一下n就可以快速得到相应的m了;
而因为随着n的增大m肯定是减少的(直觉告诉我);所以枚举结束的条件就是m< n;
然后把合法的m(能整除的)记录下来;
然后再逆序输出一遍就是n < m的情况了;
(如果出现n==m则要避免第二次再输出一次);
(n==m)的情况最多就出现一次的;
而那个x=n^2*m-(m+n)(0+n-1)*n/2+n(n+1)*(2n+1)/6
可以看到出现了n^2*m;不妨就看成n^3;
而x虽然可以很大开一个立方根里面变成10^6级的了;可以结束！

【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long

using namespace std;

const LL INF = 1e18;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);

LL ans[700000][2];
LL x,n;

void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

int main()
{
    rel(n);
    LL flag =-1,num = 0;
    for (LL i = 1;;i++)
    {
        LL fz = 6*n+i*i*i-i,temp = 3*(i*i+i),key = fz/temp;
        if (fz % temp ==0 && i <=key)
        {
            num++;
            if (i==key)
                flag = num;
            ans[num][0] = i;ans[num][1] = key;
        }
        if (key < i)
            break;
    }
    if (flag == -1)
    {
        printf("%d\n",num*2);
        for (int i = 1;i <= num;i++)
            cout << ans[i][0] << " " << ans[i][1] << endl;
        for (int i = num;i>=1;i--)
            cout << ans[i][1] << " " << ans[i][0] << endl;
    }
    else
    {
        printf("%d\n",num*2-1);
        for (int i = 1;i <= num;i++)
            cout << ans[i][0] << " " << ans[i][1] << endl;
        for (int i = num;i >=1;i--)
            if (flag==i)
                continue;
            else
                cout << ans[i][1] << " " << ans[i][0] << endl;
    }
    return 0;
}