Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4083 Accepted Submission(s): 2140
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.
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.
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define maxn 100 5 #include<cstdlib> 6 using namespace std; 7 int prime[maxn+2],rank=0; 8 bool isprime[maxn+2]; 9 10 void Prim() 11 { 12 int i,j; 13 memset(isprime,true,sizeof(isprime)); 14 15 isprime[0]=isprime[1]=false; 16 for(i=0;i*i<=maxn;i++) 17 { 18 if(isprime[i]) 19 { 20 prime[rank++]=i; 21 for(j=2*i;j<=maxn;j+=i) 22 isprime[j]=false; 23 } 24 } 25 for(j=i;j<maxn;j++) 26 if(isprime[j]) 27 prime[rank++]=j; 28 } 29 30 int main() 31 { 32 33 Prim(); 34 for(int i=0;i<rank;i++) 35 printf("%d ",prime[i]); 36 puts(""); 37 return 0; 38 }
这样处理之后会出现,需要对所需数组,进行查找,线性表中最快的查找方法为 二叉搜索....
何为二叉搜索.....
代码如下:
1 int maze[maxn+1]; 2 int two_find(int a[],int m,int n) 3 { 4 int left=0,right=n,mid; 5 while(left<right) 6 { 7 mid=(left+right)/2; 8 if(a[mid]==m) 9 break; 10 else 11 if(a[mid]<m) 12 left=mid+1; 13 else right=mid-1; 14 } 15 return mid; 16 }
剩下的就是对问题进行遍历了....由于求最大值,pq.....所以采取从最大处开始搜索......
代码:
1 #include<stdio.h> 2 #include<string.h> 3 #define maxn 100000 4 #include<stdlib.h> 5 int prime[maxn+2],step=0; 6 bool bol[maxn+5]; 7 int two_find(int m) 8 { 9 int left=0,right=step-1; 10 int mid; 11 while(left<right) 12 { 13 mid=(left+right)/2; 14 if(prime[mid]==m) 15 break; 16 else if(prime[mid]>m) 17 right=mid-1; 18 else left=mid+1; 19 } 20 return mid; 21 } 22 int main() 23 { 24 int m,a,b,i,j,k; 25 /*¿ìËÙËØÊý±í*/ 26 memset(bol,true,sizeof(bol)); 27 bol[0]=bol[1]=false; 28 prime[step++]=2; 29 /*³ýȥżÊý*/ 30 for(i=4;i<=maxn;i+=2) 31 bol[i]=false; 32 /*¿ªÊ¼*/ 33 for(i=3;i*i<=maxn;i++) 34 { 35 /*ΪËØÊý£¬´æÈ룬³ýµôÆ䱶Êý*/ 36 if(bol[i]) 37 { 38 prime[step++]=i; 39 /*´ÓÈý±¶¿ªÊ¼*/ 40 for(k=2*i,j=i*i; j<=maxn;j+=k) 41 bol[j]=false; 42 } 43 } 44 /*´òÏÂÒ»°ëËØÊý*/ 45 for( ; i<=maxn ; i++ ) 46 if(bol[i]) 47 prime[step++]=i; 48 49 while(scanf("%d%d%d",&m,&a,&b),m+a+b) 50 { 51 double cal=(double)a/b; 52 int max=two_find(m),ans=0,ansx,ansy; 53 for(i=max;i>=0;i--) 54 { 55 for(j=i;j<=max;j++) 56 { 57 if(prime[i]*prime[j]>m||(double)prime[i]/prime[j]<cal) 58 break; 59 if(prime[i]<=m/prime[j]&&(double)prime[i]/prime[j]>=cal) 60 { 61 if(ans<prime[i]*prime[j]) 62 { 63 ans=prime[i]*prime[j]; 64 ansx=i; 65 ansy=j; 66 } 67 } 68 } 69 } 70 printf("%d %d ",prime[ansx],prime[ansy]); 71 } 72 return 0; 73 }