杭电ACM 1239 简单的搜索类 Callin…

   
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K
(Java/Other)
Total Submission(s) : 6 Accepted Submission(s) : 2

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Problem Description

A message from humans to extraterrestrial intelligence was
sent through the Arecibo radio telescope in Puerto Rico on the
afternoon of Saturday November 16, 1974. The message consisted of
1679 bits and was meant to be translated to a rectangular picture
with 23 * 73 pixels. Since both 23 and 73 are prime numbers, 23 *
73 is the unique possible size of the translated rectangular
picture each edge of which is longer than 1 pixel. Of course, there
was no guarantee that the receivers would try to translate the
message to a rectangular picture. Even if they would, they might
put the pixels into the rectangle incorrectly. The senders of the
Arecibo message were optimistic.We are planning a similar project. Your task in the project is to
find the most suitable width and height of the translated
rectangular picture. The term "most suitable" is defined as
follows. An integer m greater than 4 is given. A positive fraction
a / b less than or equal to 1 is also given. The area of the
picture should not be greater than m. Both of the width and the
height of the translated picture should be prime numbers. The ratio
of the width to the height should not be less than a / b nor
greater than 1. You should maximize the area of the picture under
these constraints.In other words, you will receive an integer m and a fraction a / b.
It holds that m > 4 and 0 < a / b
< 1. You should find the pair of prime numbers p, q
such that pq <= m and a / b <= p / q
<= 1, and furthermore, the product pq takes the
maximum value among such pairs of two prime numbers. You should
report p and q as the "most suitable" width and height of the
translated picture.

Input

The input is a sequence of at most 2000 triplets of positive
integers, delimited by a space character in between. Each line
contains a single triplet. The sequence is followed by a triplet of
zeros, 0 0 0, which indicated the end of the input and should not
be treated as data to be processed.The integers of each input triplet are the integer m, the numerator
a, and the denominator b described above, in this order. You may
assume 4 < m <= 100000 and 1
<= a <= b <=
1000.

Output

The output is a sequence of pairs of positive integers. The
i-th output pair corresponds to the i-th input triplet. The
integers of each output pair are the width p and the height q
described above, in this order.Each output line contains a single pair. A space character is put
between the integers as a delimiter. No other characters should
appear in the output.

Sample Input

5 1 2
99999 999 999
1680 5 16
1970 1 1
2002 4 11
0 0 0

Sample Output

2 2
313 313
23 73
43 43
37 53

很简单的一个搜索的小题目，也很能体现搜索的内涵所在

算法思路：

题目信息：a.给定整数m,a,b(4 < m <= 100000 and1 <= a <= b <=
1000)b.需要找到两个数(不妨设为p,q)满足以下条件：p,q均为质数；p*q<=m;a/b <= p/q <= 1;c.输出所有满足以上条件的p,q中乘积最大的一对p,q典型算法：1.典型的搜索从所有可能的p,q中寻找满足条件的一对2.p,q的要求p,q均为质数,且p<=q<=100000;3.按上述思想流程应为：a.从1—100000中搜出质数b.两层循环，试遍所有的组合(p,q可能相等）c.每种组合去判断是否符合条件，如是，将p*q与当前最大值比较，判断，保存典型算法的问题：n 超时！n 从1—100000的质数运算约为1e+8,而这只是准备工作。n 因此，如不加以分析简化此题无法在规定时间内出解再分析：p,q的范围其实可在2—50000(why?)然而，这是最小的范围吗？考虑大于10000的某个质数，不妨设为Q，另一个质数为P，则：如果P<10，P/Q<0.001如果P>10，P*Q>100000而考虑到a,b的取值范围(1<=a<=b<=1000)可知min(a/b)=0.001同时，要求： p*q<=m<=100000所以无论如何质数都不能超过10000技巧：n 搜索顺序很重要。建议从大往小搜（num:质数的个数 ）for (i=num-1;i>=0;i--)for (j=i;j<=num-1;j++)……

分析源自http://hi.baidu.com/song19870626/blog/item/8c45cf02923e8380d53f7c58.html

我的代码：

C++语言:
#include<stdio.h>
int
main()
{
int
n,k,t=0,i,j;
int
num[10000]={0};      //用num[5000]来存储2~10000中的质数
int
m,a,b,p,q,max;  for(n=2;n<10000;n++)   {  k=n/2;             for(i=2;i<=k;i++)        if(n%2==0||n%i==0)break;                                     if(i>k)        {num[t]=n;t++;}   }             //进行从2~10000的质数的筛选工作      while(scanf("%d %d
%d",&m,&a,&b)&&(m!=0)&&(a!=0)&&(b!=0))  {   max=0,p=0,q=0;   for(j=t-1;j>=0;j--)   {    if(num[j]>m/2)continue;    for(i=j;i<t;i++)    {     if(num[i]*num[j]>m||a*num[i]>b*num[j]||a>b||num[i]>m/2)break;     else         if(num[i]*num[j]>max)      {       max=num[i]*num[j];       
p=num[j];       q=num[i];      }                }   }        printf("%d %d\n",p,q);       }
}