首页 > ACM题库 > UVA > UVa-107-The Cat in the Hat[数学]
2014
02-15

UVa-107-The Cat in the Hat[数学]

UVa-107-The Cat in the Hat

Background

(An homage to Theodore Seuss Geisel)

The Cat in the Hat is a nasty creature,
But the striped hat he is wearing has a rather nifty feature.

With one flick of his wrist he pops his top off.

Do you know what’s inside that Cat’s hat?
A bunch of small cats, each with its own striped hat.

Each little cat does the same as line three,
All except the littlest ones, who just say “Why me?”

Because the littlest cats have to clean all the grime,
And they’re tired of doing it time after time!

The Problem

A clever cat walks into a messy room which he needs to clean. Instead of doing the work alone, it decides to have its helper cats do the work. It keeps its (smaller) helper cats inside its hat. Each helper cat also has helper cats in its own hat, and so on. Eventually, the cats reach a smallest size. These smallest cats have no additional cats in their hats. These unfortunate smallest cats have to do the cleaning.

The number of cats inside each (non-smallest) cat’s hat is a constant, N. The height of these cats-in-a-hat is 1 / (N + 1) times the height of the cat whose hat they are in.

  • The smallest cats are of height one;
  • these are the cats that get the work done.

All heights are positive integers.

Given the height of the initial cat and the number of worker cats (of height one), find the number of cats that are not doing any work (cats of height greater than one) and also determine the sum of all the cats’ heights (the height of a stack of all cats standing one on top of another).

The Input

The input consists of a sequence of cat-in-hat specifications. Each specification is a single line consisting of two positive integers, separated by white space. The first integer is the height of the initial cat, and the second integer is the number of worker cats.

A pair of 0′s on a line indicates the end of input.

The Output

For each input line (cat-in-hat specification), print the number of cats that are not working, followed by a space, followed by the height of the stack of cats. There should be one output line for each input line other than the “0 0″ that terminates input.

Sample Input

216 125
5764801 1679616
0 0

Sample Output

31 671
335923 30275911

Analysis

算法并不复杂,主要在于浮点数精度问题的处理上。用log取对数再取整时,要注意4舍5入。从干活的猫的数量入手,因为必然是平方数,遍例所有可能的因子就能找到N的值。通过实验发现,OJ给的数据中,第一只猫的高度必然是一个等比数列之和(包括1),因此如果能够根据等比数列公式直接解出公比,即N + 1,那就非常好办了。但目前我还没有找到比较好的解法,只能先按老办法算。

Solution

#include <iostream>
#include <math.h>
using namespace std;
int main(void) {
	int nFirstHeight, nWorking;
	while (cin >> nFirstHeight >> nWorking) {
		if (nFirstHeight == 0) {
			break;
		}
		int nCatCnt = 0, nTotalHeight = 0;
		if (nWorking == 1) {
			for (int k = nFirstHeight; k >= 1; k /= 2, ++nCatCnt) {
				nTotalHeight += k;
			}
			cout << nCatCnt - 1<< ' ' << nTotalHeight << endl;
			continue;
		}
		float fLog = log((float)nWorking), fLevel;
		int nSqrt = (int)(sqrt((float)nWorking) + 0.5);
		int nLevel = 0, nHeight = nFirstHeight, nBranch = 2;
		for (; nBranch <= nSqrt; ++nBranch) {
			fLevel = fLog / log((float)nBranch);
			int nTempLevel = (int)(fLevel + 0.5);
			if (fabs((fLevel - nTempLevel)) < 1e-4f) {
				nLevel = nTempLevel + 1;
				if ((int)(pow((float)(nBranch + 1),
					nTempLevel) + 0.5) == nFirstHeight) {
						break;
				}
			}
		}
		if (nLevel == 0) {
			nLevel = 2;
			nBranch = nWorking;
		}
		for (int k = 0, nCats = 1; k < nLevel - 1; ++k) {
			nCatCnt += nCats;
			nTotalHeight += nCats * nHeight;
			nCats *= nBranch;
			nHeight /= (nBranch + 1);
		}
		cout << nCatCnt << ' ' << nTotalHeight + nWorking << endl;
	}
	return 0;
}

  1. 换句话说,A[k/2-1]不可能大于两数组合并之后的第k小值,所以我们可以将其抛弃。
    应该是,不可能小于合并后的第K小值吧

  2. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?

  3. 有一点问题。。后面动态规划的程序中
    int dp[n+1][W+1];
    会报错 提示表达式必须含有常量值。该怎么修改呢。。