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 = ;i<=10000000LL||i*i*i<=x;i++)
{
long long a = (i*i+i-(i+)*i/2LL);
long long b = (i*i+i-(i+)*i*i/2LL-(i+)*i+(i*(i+)*(*i+)/));
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\n",ans1.size());
for(int i=;i<ans1.size();i++)
{
printf("%lld %lld\n",ans1[i].x,ans1[i].y);
}
}
代码
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