poj 1595 Prime Cuts

时间：2015-02-23 08:33:33 阅读：175 评论：0 收藏：0 [点我收藏+]

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Prime Cuts

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10610		Accepted: 4046

Description

A prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In this problem you are to write a program that will cut some number of prime numbers from the list of prime numbers between (and including) 1 and N. Your program will read in a number N; determine the list of prime numbers between 1 and N; and print the C*2 prime numbers from the center of the list if there are an even number of prime numbers or (C*2)-1 prime numbers from the center of the list if there are an odd number of prime numbers in the list.

Input

Each input set will be on a line by itself and will consist of 2 numbers. The first number (1 <= N <= 1000) is the maximum number in the complete list of prime numbers between 1 and N. The second number (1 <= C <= N) defines the C*2 prime numbers to be printed from the center of the list if the length of the list is even; or the (C*2)-1 numbers to be printed from the center of the list if the length of the list is odd.

Output

For each input set, you should print the number N beginning in column 1 followed by a space, then by the number C, then by a colon (:), and then by the center numbers from the list of prime numbers as defined above. If the size of the center list exceeds the limits of the list of prime numbers between 1 and N, the list of prime numbers between 1 and N (inclusive) should be printed. Each number from the center of the list should be preceded by exactly one blank. Each line of output should be followed by a blank line. Hence, your output should follow the exact format shown in the sample output.

Sample Input

Sample Output

21 2: 5 7 11

18 2: 3 5 7 11

18 18: 1 2 3 5 7 11 13 17

100 7: 13 17 19 23 29 31 37 41 43 47 53 59 61 67

Source

South Central USA 1996

题意：给定你一个数n，让你求出1-n内有多少个素数，再给你一个数d，如果2*d大于素数的个数则全部输出；

否则，如果个数为单数，输出2*d - 1个并且以中间那个素数为中心分别向两边输出d-1个；

如果偶数个，输出2*d个，也是以中间那个素数为中心向两边扩展；————网上查的

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 #define max 1000
 6 int prime[max+10];
 7 void get_prime(int n){
 8     int i;
 9     prime[0]=1;
10     prime[1]=2;
11     int index=1;
12     for(i=3;prime[index]<=n;i++){//得到第一个大于n的素数 
13         int j=1;//j不能==0!! 
14         for(;j<=index&&prime[j]*prime[j]<=i;j++){
15             if(!(i%prime[j])){
16                 break;
17             }
18         }
19         if(prime[j]*prime[j]>i){
20             prime[++index]=i;
21         }
22     }
23     //cout<<prime[index]<<endl;
24 }
25 int main()
26 {
27     get_prime(max);
28     int n,c;
29     while(cin>>n>>c){
30         int i=0,j,s,e;
31         while(prime[i]<=n){
32             i++;
33         }
34         if(i%2){
35             if(i<2*c-1){
36                 s=0;
37                 e=i-1;
38             }
39             else{
40                 s=i/2-(c-1);
41                 e=i/2+(c-1);
42             }
43         }
44         else{
45             if(i<2*c){
46                 s=0;
47                 e=i-1;
48             }
49             else{
50                 s=i/2-1-(c-1);
51                 e=i/2+(c-1);
52             }
53         }
54         cout<<n<<" "<<c<<":";
55         for(j=s;j<=e;j++){
56             cout<<" "<<prime[j];
57         }
58         cout<<endl;
59         cout<<endl;//格式注意
60     }
61     return 0;
62 }

poj 1595 Prime Cuts

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原文地址：http://www.cnblogs.com/Deribs4/p/4297860.html

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