CD Making 逻辑数学题

最新推荐文章于 2018-07-20 15:59:30 发布

UCAS王小二

最新推荐文章于 2018-07-20 15:59:30 发布

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本文介绍了一种算法解决方案，用于解决在特定约束条件下（如避免使用数字13）将歌曲最优地分配到CD上的问题。通过分析歌曲数量与每张CD的最大容量，提出了一种高效的方法来确定所需的最少CD数量。

Tom has N songs and he would like to record them into CDs. A single CD can contain at most K songs. In addition, Tom is very superstitious and he believes the number 13 would bring bad luck, so he will never let a CD contain exactly 13 songs. Tom wants to use as few CDs as possible to record all these songs. Please help him.

Input

There are T test cases. The first line gives T, number of test cases. T lines follow, each contains N and K, number of songs Tom wants to record into CDs, and the maximum number of songs a single CD can contain.

1≤N≤1000,1≤K≤1000

Output

For each test case, output the minimum number of CDs required, if the above constraints are satisfied.

Sample Input

2
5 2
13 13

Sample Output

3
2

Hint

If Tom has 5 songs to record and a single CD can contain up to 2 songs, Tom has to use 3 CDs at minimum, each contains 2, 2, 1 songs, respectively.

In the second case, Tom wants to record 13 songs, and a single CD can hold 13 songs at most. He would have been able to use only 1 CD if he were not so superstitious. However, since he will not record exactly 13 songs into a single CD, he has to use 2 CDs at least, the first contains 12 songs and the second contains one(Other solutions to achieve 2 CDs are possible, such as (11, 2), (10, 3), etc.).

题目大意:

把n首歌放到CD中，每个CD最多放k首歌，但是每个CD不能放13首，给出n和k求需要的CD数；

思路：k=13时，把k--；k<13时，相除看余数是否进一；k>13时，看余数是否为13，考虑n/k是否为0；

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#define max_n 10010
typedef long long LL;
using namespace std;

int main()
{
	int t,n,k;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&n,&k);
		if(k==13)
		{
			k--;
			if(n%k==0)
				printf("%d\n",n/k);
			else printf("%d\n",n/k+1);
		}
		else if(n==13 && n<k) //特殊考虑 
			printf("2\n");
		else
		{
			if(n%k==0)
				printf("%d\n",n/k);
			else if(n%k==13)
			{
				if(k==14)
					printf("%d\n",n/k+2);
				else printf("%d\n",n/k+1);
			}
			else printf("%d\n",n/k+1);
		}
	}
	return 0;
}