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
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.
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.
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.
5 1 2 99999 999 999 1680 5 16 1970 1 1 2002 4 11 0 0 0
2 2 313 313 23 73 43 43 37 53
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 0xfffffff int 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; }