Codeforces Round #332 (Div. 2)D. Spongebob and Squares 数学
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.
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.
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.
26
6
1 26
2 9
3 5
5 3
9 2
26 1
2
2
1 2
2 1
8
4
1 8
2 3
3 2
8 1
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
///
#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=,f=;char ch=getchar();
while(ch<''||ch>''){
if(ch=='-')f=-;ch=getchar();
}
while(ch>=''&&ch<=''){
x=x*+ch-'';ch=getchar();
}return x*f;
}
//****************************************
const int N=+;
#define maxn 100000+5 ll x; int main()
{
x=read();
vector<pair<ll, ll> > ans;
ll sum=;
for(ll i=; sum<=x; i++)
{
sum+=1ll*i*i;
long long d=x-sum;
long long k=1LL*i*(i+)/;
if(d%k==)
{
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=;i<ans.size();i++)
printf("%I64d %I64d\n", ans[i].first, ans[i].second);
return ;
}
代码
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