codeforces 599D. Spongebob and Squares【推公式+暴力枚举】_spongebob is already tired trying to reason his we-优快云博客

本文链接：https://blog.youkuaiyun.com/Bcwan_/article/details/52372323

本文介绍了一道算法题目，任务是找出所有可能的表格尺寸（n行m列），使得这样的表格中恰好包含x个不同的正方形。文章提供了一个C++实现方案，通过枚举方法来解决该问题，并给出了具体的代码实现。

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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 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.

   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.

$\text{[math]}$

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

$\text{[math]}$

把③化简得到：6*x = 6*n*n*m - 3*(n+m)*n*(n-1) + n*(n-1)*(2*n-1)

进而有 (3*n*n+3*n)*m = 6*x + n*n*n - n

枚举n求m即可

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
typedef long long ll;
using namespace std;
struct Node{
    ll nn,mm;
};
vector<Node>ans;
ll high=5000000;
int main()
{
    ll x;
    scanf("%I64d",&x);
    for(ll n=1;n<=high;n++){  //枚举n
        ll a=6*x+n*n*n-n;
        ll b=3*n*n+3*n;
        if(a%b!=0)continue;
        ll m=a/b;
        if(n>m)break;         //枚举到n<=m即可，因为在n>m时的结果等于n<m所有结果(n,m)的交换,即(m,n)
        Node now;
        now.nn=n;
        now.mm=m;
        ans.push_back(now);
    }
    int top=ans.size()-1;
    //如果最末答案为(n,n),则总数为2*(n<=m的对数)-1,否则总数为2*(n<=m的对数)
    if(ans[top].nn==ans[top].mm) top=ans.size()*2-1;
    else top=ans.size()*2;
    printf("%d\n",top);
    int i;
    top=ans.size()-1;
    for(i=0;i<=top;i++){
        printf("%I64d %I64d\n",ans[i].nn,ans[i].mm);
    }
    if(ans[top].nn==ans[top].mm)i=top-1;
    else  i=top;
    for(;i>=0;i--){
        printf("%I64d %I64d\n",ans[i].mm,ans[i].nn);
    }
    return 0;
}