【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>>
【问题描述】
Twin primes are pairs of primes of the form (p, p+2). The term "twin prime" was coined by Paul Stäckel (1892-1919). The first few twin primes are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43). In this problem you are asked to find out the S-th twin prime pair where S is an integer that will be given in the input.
Input
The input will contain less than 10001 lines of input. Each line contains an integers S (1<=S<=100000), which is the serial number of a twin prime pair. Input file is terminated by end of file.
Output
For each line of input you will have to produce one line of output which contains the S-th twin prime pair. The pair is printed in the form (p1,<space>p2). Here <space> means the space character (ASCII 32). You can safely assume that the primes in the 100000-th twin prime pair are less than 20000000.
Sample Input
1
2
3
4
Sample Output
(3, 5)
(5, 7)
(11, 13)
(17, 19)
【解题思路】
先离线计算素数筛,再筛选出双素数保存较小的一个,判断输出。
【具体实现】
#include<iostream>
#define N 20000000
#define n 100000
#define Inp 10001
using namespace std;
int SIGN[N] = { 0 };
int ANS[n] = { 0 };
int main(){
/*素数筛*/
SIGN[0] = SIGN[1] = 1;
for (int i = 2; i < N;++i)
if (!SIGN[i])
for (int j = i * 2; j < N; j += i)
SIGN[j] = 1;
/*若为双素数,保存较小的*/
int s=0;
for (int k = 3; k < N;++k)
if (SIGN[k] == 0 && SIGN[k + 2] == 0)
ANS[++s] = k;
int num;
int times = 0;
while (cin >> num)
if (num >= 1 && num <= n&× < Inp)
cout << '(' << ANS[num] << ',' << ' ' << ANS[num] + 2 << ')' << endl;
return 0;
}
【额外补充】
1.int型数组
函数体外定义,不初始化会被默认初始化为0;
函数体外定义,初始化第一个,后面的都会默认初始化为0;
函数体外定义,初始化前一部分,后面的都会默认初始化为0;
函数体内定义,不初始化会是垃圾值。
函数体内定义,初始化第一个,后面的都会默认初始化为0;
函数体内定义,初始化前一部分,后面的都会默认初始化为0;
2.float型数组
函数体外定义,不初始化会被默认初始化为0;
函数体外定义,初始化第一个,后面的都会默认初始化为0;
函数体外定义,初始化前一部分,后面的都会默认初始化为0;
函数体内定义,不初始化会是垃圾值。
函数体内定义,初始化第一个,后面的都会默认初始化为0;
函数体内定义,初始化前一部分,后面的都会默认初始化为0;
3.char型数组
函数体外定义,不初始化会被默认初始化为空;
函数体外定义,初始化第一个,后面的都会默认初始化为空;
函数体外定义,初始化前一部分,后面的都会默认初始化为空;
函数体内定义,不初始化会被默认初始化为空;
函数体内定义,初始化第一个,后面的都会默认初始化为空;
函数体内定义,初始化前一部分,后面的都会默认初始化为空;
来源:oschina
链接:https://my.oschina.net/u/2725918/blog/664827