codechef Ciel and Receipt题解

Tomya is a girl. She loves Chef Ciel very much.

Tomya like a positive integer p, and now she wants to get a receipt of Ciel's restaurant whose total price is exactly p.
The current menus of Ciel's restaurant are shown the following table.

Name of Menu	price
eel flavored water	1
deep-fried eel bones	2
clear soup made with eel livers	4
grilled eel livers served with grated radish	8
savory egg custard with eel	16
eel fried rice (S)	32
eel fried rice (L)	64
grilled eel wrapped in cooked egg	128
eel curry rice	256
grilled eel over rice	512
deluxe grilled eel over rice	1024
eel full-course	2048

Note that the i-th menu has the price 2^i-1 (1 ≤ i ≤ 12).

Since Tomya is a pretty girl, she cannot eat a lot.
So please find the minimum number of menus whose total price is exactly p.
Note that if she orders the same menu twice, then it is considered as two menus are ordered. (SeeExplanations for details)

Input

The first line contains an integer T, the number of test cases.
Then T test cases follow.
Each test case contains an integer p.

Output

For each test case, print the minimum number of menus whose total price is exactly p.

Constraints

1 ≤ T ≤ 5
1 ≤ p ≤ 100000 (10⁵)
There exists combinations of menus whose total price is exactly p.

Sample Input

4
10
256
255
4096

Sample Output

2
1
8
2

本题因为数据特殊，原来能够使用贪心法的。

时间效率能够达到O(1)。由于最后有个1的数组能被不论什么整数整除，所以是必定有解的。

#include <stdio.h>

int CielandReceiptCal()
{
	const static int MENUS_NUM = 12;

	int T;
	scanf("%d", &T);
	while (T--)
	{
		int P;
		scanf("%d", &P);
		int largest = 2048, ans = 0;
		for (int i = MENUS_NUM - 1; i >= 0 ; i--)
		{
			ans += P / largest;
			P %= largest;
			largest >>= 1;
		}
		printf("%d\n", ans);
	}
	return 0;
}

当然无数据特殊性的时候，就要使用dp了：

int CielandReceiptDP()
{
	const static int MENUS_NUM = 12;
	const static int MENUS[MENUS_NUM] = {1, 2, 4, 8, 16, 32, 64, 
		128, 256, 512, 1024, 2048};

	int T;
	scanf("%d", &T);
	while (T--)
	{
		int P;
		scanf("%d", &P);
		int *tbl = new int[P+1];//这里的[]写成()错误
		tbl[0] = 0;
		for (int i = 1; i <= MENUS_NUM; i++)
		{
			for (int j = MENUS[i-1]; j <= P; j++)
			{
				tbl[j] = tbl[j-MENUS[i-1]] + 1;
			}//处理4096000上百万个数据也能秒杀
		}
		printf("%d\n", tbl[P]);
		delete [] tbl;
	}
	return 0;
}