Codeforces Round #332 (Div. 2) D. Spongebob and Squares 数学题枚举

D. Spongebob and Squares

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/599/problem/D

Description

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 Input

26

Sample Output

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

HINT

题意

给你x，然后让你找有多少个n*m的矩形，可以由x个相同的多边形组成

题解：

数学题，这道题实际上是问，f(n,m) = sigma(k=1,k=min(n,m))(n-k+1)*(m-k+1)=x的解有多少个

化简之后，我们可以得到f(n,m) = n^2m+n^2+n*m+n-(n+1)*n/2*(n+m+2)+n*(n+1)*(2n+1)/6

这个式子是一个关于m的一次函数，我们枚举n就好了

就可以求m了，注意break条件

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;

struct node
{
    long long x,y;
};
bool cmp(node a,node b)
{
    return a.y>b.y;
}
vector<node> ans1;
int main()
{
    long long x;cin>>x;
    for(long long i = 1;i<=10000000LL||i*i*i<=x;i++)
    {
        long long a = (i*i+i-(i+1)*i/2LL);
        long long b = (i*i+i-(i+1)*i*i/2LL-(i+1)*i+(i*(i+1)*(2*i+1)/6));
        long long y = x;
        long long t= (y-b)/a;
        if(i>t)continue;
        if(a*t+b==y)
        {
            if(i==t)
            {
                node k;k.x = i,k.y = t;
                ans1.push_back(k);
            }
            else
            {
                node k;k.x = i,k.y = t;
                ans1.push_back(k);
                k.x = t,k.y = i;
                ans1.push_back(k);
            }
        }
    }
    sort(ans1.begin(),ans1.end(),cmp);
    printf("%d
",ans1.size());
    for(int i=0;i<ans1.size();i++)
    {
        printf("%lld %lld
",ans1[i].x,ans1[i].y);
    }
}

代码