Codeforces 599D Spongebob and Squares【思维枚举+数学方程】

时间:2022-07-28 17:07:47

D. Spongebob and Squarestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard 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 exactlyx distinct squares in the table consisting ofn rows and m columns. For example, in a3 × 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 exactlyx distinct squares inside.

Then print k pairs of integers describing the tables. Print the pairs in the order of increasingn, and in case of equality — in the order of increasingm.

ExamplesInput
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.

Codeforces 599D Spongebob and Squares【思维枚举+数学方程】

In a 2 × 3 table there are 6 1 × 1 squares and 22 × 2 squares. That is equal to 8 squares in total.

Codeforces 599D Spongebob and Squares【思维枚举+数学方程】


题目大意:

让你求一共有多少种方案,使得n*m的正方形中,有k个子正方形。


思路:


1、首先很明显,对于一个3*5的正方形中,其有子正方形:n*m+(n-1)*(m-1)+(n-2)*(m-2)==26个;

那么同理,对于一个n*m的正方形中,其有子正方形:

tmp=min(n,m);

数量=n*m*tmp-【(tmp)*(tmp-1)/2】*(x+y)+1^2+2^2+3^2+4^2+..........(tmp-1)^2;

那么问题就转化成为:有多少个x,使得n*m*tmp-【(tmp)*(tmp-1)/2】*(n+m)+1^2+2^2+3^2+4^2+..........(tmp-1)^2==X.


2、接下来观察等式,等式右侧一定是大于等于1的一个数,那么肯定一点,如果1^2+2^2+..........+(tmp-1)^2大于了X.那么显然就是无解的。

那么我们着重讨论1^2+2^2+.............+(tmp-1)^2.

打个暴力的表,并且存到sum【i】,表示1^2+2^2+.........+i^2的值。

当sum【i】>1e18的时候跳出,不难发现,这个数组的长度是1e6+的。

那么我们可以考虑枚举n,同时设定tmp==n,对于剩下已知部分,很容易就能求得m。

接下来的任务就变成了解方程式,求y.

对于求出的解进行统计即可。


3、注意n==m的时候,千万不能多输出。(被这里坑了一波,很蓝廋);


Ac代码:

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define ll __int64
struct node
{
ll x,y;
}ans[1442275];
ll sum[1442275];
int tot;
int cmp(node a,node b)
{
return a.x<b.x;
}
void init()
{
ll summ=0;
for(ll i=1;i<=100000000;i++)
{
summ+=i*i;
if(summ>=1000000000000000000)
{
tot=i-1;
break;
}
sum[i]=summ;
}
}
int main()
{
init();
ll x;
while(~scanf("%I64d",&x))
{
int cnt=0;
for(ll i=1;i<=tot;i++)
{
if(x>=sum[i])
{
ll right=x-sum[i-1];
right+=(i-1)*i*i/2;
right*=2;
if(right%(i*i+i)==0)
{
ll y=right/(i*i+i);
ans[cnt].x=i;
ans[cnt++].y=y;
ans[cnt].x=y;
ans[cnt++].y=i;
if(i==y)cnt--;
}
}
}
sort(ans,ans+cnt,cmp);
printf("%d\n",cnt);
for(int i=0;i<cnt;i++)
{
printf("%I64d %I64d\n",ans[i].x,ans[i].y);
}
}
}