PAT (Advanced Level) Practice 1103

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PAT (Advanced Level) Practice 1103

原题链接

1. 题目大意

给出三个正数 N , P , K N,P,K N,P,K,其中范围分别是
N ≤ 400 , K ≤ N , 1 < P ≤ 7 N\leq400,K\leq{N},1<P\leq7 N≤400,K≤N,1<P≤7
找出满足下列条件的序列, n [ 1 ] … n [ K ] n[1]…n[K] n[1]…n[K]

  • N = n [ 1 ] P + . . . n [ K ] P N = n[1]^P + ... n[K]^P N=n[1]P+...n[K]P
  • 多组解保证 m a x ( n [ 1 ] + … n [ k ] ) max(n[1]+…n[k]) max(n[1]+…n[k])
  • 若 m a x ( n [ 1 ] + … n [ k ] ) max(n[1]+…n[k]) max(n[1]+…n[k])相同则保证序列是最大字典序的

2. 解法

我们通过题目给出的样例思考,169,5,2,首位最大的数字应该是12,怎么算出来的呢?因为5个坑位,不能有0,题目要求保证序列最后是最大字典序的,所以我们可以规定序列是一个单调不增的,这样又限制了第一个坑位的下界。比如169,5,2,需要给后面四个留数值,所以 a [ 1 ] 2 < = 169 − 4 a[1]^2<=169-4 a[1]2<=169−4,并且 5 × a [ 1 ] 2 > = 169 5{\times}a[1]^2>=169 5×a[1]2>=169,如此一来, a [ 1 ] a[1] a[1]的范围就被限制在了6和12之间。题目给出的P的范围是2到7,实际上情况不是很多,可以考虑使用dfs枚举所有情况。

如何确定dfs的范围呢?
我们定义函数dfs(depth,remain,thread),其中,depth代表dfs进行到了序列的第几位,remain表示n减去之前序列的P次方剩余的数量,thread表示上界,由于我们枚举的是不上升序列。

对于上界,由于当前遍历到的深度为depth,即在当前位之后还有k-depth位,对于这k-depth个坑位至少每个要留下个1,即k-depth个 1 p 1^p 1p,所以要找到上界 u p up up使得

u p p < = n ( u p + 1 ) p > n up^p<=n \quad (up+1)^p>n upp<=n(up+1)p>n

对于下界,我原先的代码设置为了1,但是这样会导致一些极端数据严重超时,比如400 200 2。究其原因,下界设置为1也有很多不必要的搜索过程。于是乎考虑dfs的剪枝。由于我们需要给出的是不上升的序列,我们可以增加如下剪枝

if ((k-depth+1)*pow(thread,p)<remain) return;

thread表示上界,即如果之后所有的坑位都使用最大上界依然无法填补未达到的remain之后返回,增加了这个剪枝之后唯一超时的点跑过去了。

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1000

int n,k,p;
int a[MAXN];
int ans[MAXN],maxx;
int re;

int findup(int n,int p)
//寻找上界 
{
	int x=0;
	while (pow(x,p)<=n)
	{
		if ((pow(x,p)<=n)&&pow(x+1,p)>n) return x;
		x++;
	}
}

int findlow(int n,int k,int p)
//寻找首位下界 
{
	int x=0;
	while (k*pow(x,p)<n)
	{
		if ((k*pow(x,p)<n)&&(k*pow(x+1,p)>=n)) return x+1;
		x++;
	}
}

void dfs(int depth,int remain,int thread)
{
	if ((k-depth+1)*pow(thread,p)<remain) return;//必要的剪枝 
	if (depth==k)
	{
		//达到搜索深度了 
		for (int i=thread;i>=1;i--)
			//找到pow(i,p)刚好填补remain 
			if (remain==pow(i,p))
			{
				a[depth]=i;
				int tot=0;
				for (int i=1;i<=k;i++) tot+=a[i];
//				for (int i=1;i<=k;i++) printf("%d ",a[i]);
//				cout<<endl;
				if (tot>maxx)
				{
					maxx=tot;
					for (int i=1;i<=k;i++) ans[i]=a[i];
				}
				return;
			}
		return;
	}
	int y=k-depth;
	int x=remain-y;
	int up=min(thread,findup(x,p));//取计算出的上界和在这坑位之前的上界的最小值 
	for (int i=up;i>=1;i--)
	{
		a[depth]=i;
		int re=remain-pow(i,p);
		//i作为新的上界,继续搜索 
		dfs(depth+1,re,i);
	}
}

int main()
{
	scanf("%d%d%d",&n,&k,&p);
	if (k==1)
	{
		int x=0;
		while (pow(x,p)<n) x++;
		if (pow(x,p)==n) 
		{
			printf("%d = ",n);
			printf("%d^%d",x,p);
		}
		else printf("Impossible");
		return 0;
	}
	int x=n-(k-1);
	int up=findup(x,p);
	int low=findlow(n,k,p);
	for (int i=up;i>=low;i--)
	{
		re=n;
		a[1]=i;
		re-=pow(i,p);
		dfs(2,re,i);
	}
	if (ans[1])
	{
		printf("%d = ",n);
		for (int i=1;i<k;i++)
			printf("%d^%d + ",ans[i],p);
		printf("%d^%d",ans[k],p);
	}
	else printf("Impossible");
	return 0;
}
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关于参照《算法笔记》学习DFS时的一些自身理解 首先DFS是一种“出栈入栈”思想。 到达分岔口时入栈,离开分岔口(即该分岔口处所有可走路径都已遍历完成)或者死胡同时出栈。 这里附上我觉得对我理解帮助比较大的一篇博客 递归是用来解决该类问题的有利工具 毕竟递归即是系统利用系统栈来存放递归的每一层状态。因此其本质还是栈。 在使用递归时,递归式即为分岔口,递归边界即为死胡同。 拿Fibonacci数列举...

tan_1999的博客 362

RU DOS版、EFI版32bit/64bit BIOS常用工具

DOS版、EFI版的RU工具,可用来查看IO、MEM、PCI、SMBIOS等资源。亲测在EFI SHELL下可用。

PAT (Advanced Level) Practice 1101~1120

文章目录11011102110311041105110611071108110911101111111211131114111511161117111811191120 1101 1102 1103 1104 一道纯考察数学规律的题目。 第 i 个数 a b c d 在长度为1的片段中出现的次数 1 1 1 1 在长度为2的片段中出现的次数 1 2 2 1 在长度为3的片段中出现的次数 1 2 2 1 在长度为4的片段中出现的次数 1 1 1 1 总计出现次数 4 6

张小艾博客 959

PAT (Basic Level) Practice (中文)(参考答案)1061-1120

PAT (Basic Level) Practice (中文)(参考答案)1061-1120

少侠PSY的博客 1040

PAT (Advanced Level) Practice 1098 Insertion or Heap Sort(25分)

According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, an

一只小羊 253

PAT (Advanced Level) Practice 1089 Insert or Merge(25分)【排序】

According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, an

一只小羊 279

PAT (Advanced Level) Practice 1089 Insert or Merge (25分)

1.题目 According to Wikipedia: Insertion sortiterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input...

且慢 142

【PAT (Advanced Level) Practice】1089 Insert or Merge (25 分)

According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, an

多思考🤔~ 436

PAT (Advanced Level) Practice 1089 Insert or Merge (25 分)

1089 Insert or Merge (25 分) According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one e...

就当一次路过丶的沙滩 216

PAT (Advanced Level) Practice 1098 Insertion or Heap Sort (25 分)

Insertion or Heap Sort According to Wikipedia:Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element...

就当一次路过丶的沙滩 386

【PAT (Advanced Level) Practice】1098 Insertion or Heap Sort (25 分)

According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, an

多思考🤔~ 366

PAT (Advanced Level) Practice_1089 Insert or Merge (25分)(C++_排序)

According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, an

ccql's Blog 364
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