hdu1239 Calling Extraterrestrial Intelligence Again (枚举)

Calling Extraterrestrial Intelligence Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3929    Accepted Submission(s): 2062

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 299999 999 9991680 5 161970 1 12002 4 110 0 0

 

Sample Output

2 2313 31323 7343 4337 53

 

Source

Asia 2002, Kanazawa (Japan)

 

大意：

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

分析：

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;

  易知 p,q [2,50000]; //根据素数定理：n/ln(n)近视为不超过n的素数的个数。n=10^4,n/ln(n)  = 1229

  现证明p,q不超过10000:

  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。（事实上，不会超过9091）

  

code:

#include <stdio.h>#include <string.h>#define N 10000#define INF 0xfffffffint prime[N],cnt;bool mark[N];void sieve() {    int i,j;    memset(mark,0,sizeof(mark));    cnt = 0,prime[cnt++] = 2;    for(i=3; i<N; i+=2) {        if(!mark[i]) prime[cnt++] = i;        for(j=1; j<cnt&&prime[j]*i<N; ++j) {            mark[i*prime[j]] = 1;            if(!(i%prime[j])) break;        }    }}int main() {    int m ,a ,b, i, j;    double s;    int sub,tmp;    sieve();    while(scanf("%d%d%d",&m,&a,&b),a+b+m) {        s  = (double)a/b;        sub = -INF;        for(i=cnt-1; i>=0; i--)            for(j=i; j<cnt; j++) {                if(prime[j]>m||prime[j]*prime[i]>m||(double)prime[i]/prime[j]<s)                    continue;                tmp = prime[i]*prime[j];                if(tmp>sub) {                    sub=tmp;                    a= prime[i];                    b=prime[j];                }            }        printf("%d %d\n",a,b);    }    return 0;}