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

D. Spongebob and Squares

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

61 262 93 55 39 226 1

input

2

output

21 22 1

input

8

output

41 82 33 28 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：给你x，问你多少种n*m的情况使得，在当前这个矩形内小正方形的个数为x

题解：

列式推： sigma(k=1,k=min(n,m))(n-k+1)*(m-k+1)=x；

我们枚举n，得到m

///1085422276
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back

inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){
        if(ch==‘-‘)f=-1;ch=getchar();
    }
    while(ch>=‘0‘&&ch<=‘9‘){
        x=x*10+ch-‘0‘;ch=getchar();
    }return x*f;
}
//****************************************
const int N=100000+350;
#define maxn 100000+5

ll x;

int main()
{
    x=read();
    vector<pair<ll,  ll> >  ans;
    ll sum=0;
    for(ll i=1; sum<=x; i++)
    {
        sum+=1ll*i*i;
        long long d=x-sum;
        long long k=1LL*i*(i+1)/2;
        if(d%k==0)
        {
            ans.push_back({i, d/k+i});
            ans.push_back({d/k+i, i});
        }
    }
    sort(ans.begin(), ans.end());
    ans.resize(unique(ans.begin(), ans.end())-ans.begin());
    printf("%d\n", ans.size());
    for(ll i=0;i<ans.size();i++)
        printf("%I64d %I64d\n", ans[i].first, ans[i].second);
    return 0;
}

代码