Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

A message from humans to extraterrestrial inteIigence was sent through the Arecibo radio telescope in Puerto Rico on the afternoon of Saturday November l6, l974. The message consisted of l679 bits and was meant to be translated to a rectangular picture with 23 x 73 pixels. Since both 23 and 73 are prime numbers, 23 x 73 is the unique possible size of the translated rectangular picture each edge of which is longer than l 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 haction a/b . It holds that m > 4 and 0 < a/b1 . You should find the pair of prime numbers p, q such that pqm and a/bp/q1 , 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 indicates 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 < m100000 and 1ab1000 .

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

---

題目描述：

找到兩個質數 p, q，且 pq <= m, 且符合 a/b <= p/q <= 1。

而 p*q 越大越好。

題目解法：

m <= 100000，直接窮舉即可。

#include <stdio.h>
int p[10000], pt = 0;
void sieve() {
    int mark[100000] = {};
    int i, j;
    for(i = 2; i < 100000; i++) {
        if(mark[i] == 0) {
            p[pt++] = i;
            for(j = i+i; j < 100000; j += i)
                mark[j] = 1;
        }
    }
}
int main() {
    sieve();
    int n, a, b, i, j;
    while(scanf("%d %d %d", &n, &a, &b) == 3 && n+a+b) {
        int pp, qq, mx = 0;
        for(i = 0; i < pt && p[i]*p[i] <= n; i++) {
            for(j = i; j < pt; j++) {
                if(a*p[j] > b*p[i])// a/b > p/q
                    break;
                if(p[i]*p[j] > n)   break;
                if(p[i]*p[j] > mx)
                    pp = p[i], qq = p[j], mx = pp*qq;
            }
        }
        printf("%d %d\n", pp, qq);
    }
    return 0;
}