day3 CodeForces - 472A

One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.

For example, there is a statement called the “Goldbach’s conjecture”. It says: “each even number no less than four can be expressed as the sum of two primes”. Let’s modify it. How about a statement like that: “each integer no less than 12 can be expressed as the sum of two composite numbers.” Not like the Goldbach’s conjecture, I can prove this theorem.

You are given an integer n no less than 12, express it as a sum of two composite numbers.

Input
The only line contains an integer n (12 ≤ n ≤ 106).

Output
Output two composite integers x and y (1 < x, y < n) such that x + y = n. If there are multiple solutions, you can output any of them.

Examples
Input
12
Output
4 8
Input
15
Output
6 9
Input
23
Output
8 15
Input
1000000
Output
500000 500000
Note
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output “6 6” or “8 4” as well.

In the second example, 15 = 6 + 9. Note that you can’t output “1 14” because 1 is not a composite number.
分析：题目大意就是给出不超过30种类型的方体，如何叠高才能使高度最大，并且在下面的立方体要留有少量面积给小猴子爬；
动态规划：30种类型，可以摆出180种立方体；dp[i]表示i个方体的最大高度
然后用dp[i]来存，到第i个木块，最高能够累多高。
当然。长方体先要以长度排序，长度同样则宽度小的在上

#include<iostream>
using namespace std;
#include<iomanip>
#include<queue>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
	int l,w,h;
}cd[182];
int dp[182];
bool cmp(node cd1,node cd2)
{
	if(cd1.l==cd2.l)
		return cd1.w<cd2.w;
	return cd1.l<cd2.l;
}
int main()
{
	int n;int ww=1;
	while(cin>>n,n)
	{
		memset(dp,0,sizeof(dp));
	int l,w,h;
	int t=1;
	for(int i=1;i<=n;i++)
	{
		cin>>l>>w>>h;
		cd[t].l=l;cd[t].w=w;cd[t++].h=h;
		cd[t].l=w;cd[t].w=l;cd[t++].h=h;
		cd[t].l=h;cd[t].w=l;cd[t++].h=w;
		cd[t].l=h;cd[t].w=w;cd[t++].h=l;
		cd[t].l=l;cd[t].w=h;cd[t++].h=w;
		cd[t].l=w;cd[t].w=h;cd[t++].h=l;
	}
	
	sort(cd,cd+t,cmp);
	dp[1]=cd[1].h;
	int maah;
	for(int i=2;i<t;i++)
	{
		maah=0;
		for(int j=1;j<=i;j++)
		{
			if(cd[j].l<cd[i].l&&cd[j].w<cd[i].w)
			{
				maah=max(dp[j],maah);
			}
		}
		dp[i]=cd[i].h+maah;
	};
	 maah=0;
	for(int i=1;i<t;i++)
	{
		if(maah<dp[i])
			maah=dp[i];
	}
	cout<<"Case "<<ww++<<": maximum height = "<<maah<<endl;
}
}