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 wassent through the Arecibo radio telescope in Puerto Rico on theafternoon of Saturday November 16, 1974. The message consisted of1679 bits and was meant to be translated to a rectangular picturewith 23 * 73 pixels. Since both 23 and 73 are prime numbers, 23 *73 is the unique possible size of the translated rectangularpicture each edge of which is longer than 1 pixel. Of course, therewas no guarantee that the receivers would try to translate themessage to a rectangular picture. Even if they would, they mightput the pixels into the rectangle incorrectly. The senders of theArecibo message were optimistic.
We are planning a similar project. Your task in the project is tofind the most suitable width and height of the translatedrectangular picture. The term "most suitable" is defined asfollows. An integer m greater than 4 is given. A positive fractiona / b less than or equal to 1 is also given. The area of thepicture should not be greater than m. Both of the width and theheight of the translated picture should be prime numbers. The ratioof the width to the height should not be less than a / b norgreater than 1. You should maximize the area of the picture underthese 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, qsuch that pq <= m and a / b <= p / q<= 1, and furthermore, the product pq takes themaximum value among such pairs of two prime numbers. You shouldreport p and q as the "most suitable" width and height of thetranslated picture.

Input

The input is a sequence of at most 2000 triplets of positiveintegers, delimited by a space character in between. Each linecontains a single triplet. The sequence is followed by a triplet ofzeros, 0 0 0, which indicated the end of the input and should notbe treated as data to be processed.

The integers of each input triplet are the integer m, the numeratora, and the denominator b described above, in this order. You mayassume 4 < m <= 100000 and 1<= a <= b <=1000.

Output

The output is a sequence of pairs of positive integers. Thei-th output pair corresponds to the i-th input triplet. Theintegers of each output pair are the width p and the height qdescribed above, in this order.

Each output line contains a single pair. A space character is putbetween the integers as a delimiter. No other characters shouldappear in the output.

Sample Input

5 1 299999 999 9991680 5 161970 1 12002 4 110 0 0

Sample Output

2 2313 31323 7343 4337 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 and

1 <= 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>
intmain()
{
intn,k,t=0,i,j;
intnum[10000]={0};      //用num[5000]来存储2~10000中的质数
intm,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);
       }
}