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