GPCPC_01_An_Easy_Problem

admin管理员组

文章数量:1558087

题目描述

Measuring the area of a certain shape is an important part of certain geometric problems.

For a point A(x,y) on the plane, we define the F(x,y) as ∣x∣∗∣y∣,which means the area of the rectangle with the vertex(x,0),(0,y),(0,0),(x,y).

You are given n∗m vertex with integral coordinates in the form of (i,j),and 1≤i≤n,1≤j≤m. You need to find the K-th biggest value of F(i,j),when 1≤i≤n,1≤j≤m.

It’s guranteed that K≤n∗m.

输入描述:

Three integers n,m,Kn,m,K_{}n,m,K(1≤n,m,K≤1061\leq n,m,K\leq 10^61≤n,m,K≤106). 

输出描述:

One integer,the K-th biggest value of F(i,j)F(i,j)_{}F(i,j).

示例1

输入

3 3 4

输出

4

说明

In this eaxmple, the values of all F(i,j) are 1,2,2,3,3,4,6,6,9.

So the 4-th biggest value of F(i,j)F(i,j)_{}F(i,j) is 4.

题目大意:

输入两个数n,m,k.得到从1-n 分别乘以 1-m的数.然后排列输出第K大的数

解题思路:

声明:这是看了别的大佬的做法后自己仿造的.如有侵犯请联系我删除.

利用set容器中pair中一个记录最大值,另外一个是产生最大值的数.然后最大值减去这个数再插入回容器.set容器根据值的大小排序,所以顶上一直是最大值,减一次就会产生第二大的数.以此类推可以产生排在第几的数.

限制

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

代码:

#include<iostream>
#include<set>
#include<utility>
#include<vector>
using namespace std;

typedef long long ll;

typedef pair < ll, ll> pll;

set<pll> pg;

int main()
{
	int n, m, k;
	cin >> n >> m >> k;
	for (int i = 0; i <= n; ++i)
	{
		pg.insert(pg.begin(),{ i * m,i });
	}
	k--;
	while (k--)
	{
		set<pll>::iterator it = pg.end();
		it--;
		ll val = it->first;
		ll line = it->second;

		pg.insert(pg.end(),{ val - line,line });
		pg.erase(it);
	}
	set<pll>::iterator it = --pg.end();
	cout << it->first << endl;

	return 0;
}

运行结果

本文标签： GPCPC01AnEasyProblem