USACO section 1.5 Superprime Rib(深搜)

本文深入探讨了技术领域的超级质数与超质数肋骨的概念,通过递归搜索算法构建长度为n的超质数肋骨,并详细解释了算法背后的数学原理与实现细节。

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Superprime Rib

Butchering Farmer John's cows always yields the best prime rib. You can tell prime ribs by looking at the digits lovingly stamped across them, one by one, by FJ and the USDA. Farmer John ensures that a purchaser of his prime ribs gets really prime ribs because when sliced from the right, the numbers on the ribs continue to stay prime right down to the last rib, e.g.:

     7  3  3  1

The set of ribs denoted by 7331 is prime; the three ribs 733 are prime; the two ribs 73 are prime, and, of course, the last rib, 7, is prime. The number 7331 is called a superprime of length 4.

Write a program that accepts a number N 1 <=N<=8 of ribs and prints all the superprimes of that length.

The number 1 (by itself) is not a prime number.

PROGRAM NAME: sprime

INPUT FORMAT

A single line with the number N.

SAMPLE INPUT (file sprime.in)

4

OUTPUT FORMAT

The superprime ribs of length N, printed in ascending order one per line.

SAMPLE OUTPUT (file sprime.out)

2333
2339
2393
2399
2939
3119
3137
3733
3739
3793
3797
5939
7193
7331
7333
7393

 
 
 
 
/*
ID:nealgav1
PROG:sprime
LANG:C++
*/
#include<cstdio>
#include<cmath>
int m;
bool isprim(int num)//判断质数
{
  if(num==2)return 1;
  if(num&1)
  {
    int n=(int)sqrt(num);
    for(int i=3;i<=n;i+=2)
    if(!(num%i))
    return 0;
    return 1;
  }
  return 0;
}
void dfs(int num,int n)
{
  int ans;
  if(m==n)
  {
    printf("%d\n",num);
    return;
  }
  for(int i=1;i<=9;i+=2)
  {
    ans=num*10+i;
    if(isprim(ans))
    dfs(ans,n+1);
  }
}
int main()
{
  freopen("sprime.out","w",stdout);
  freopen("sprime.in","r",stdin);
  scanf("%d",&m);
  dfs(2,1);dfs(3,1);dfs(5,1);dfs(7,1);
}
 
 
 
 
 
 
USER: Neal Gavin Gavin [nealgav1]
TASK: sprime
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3344 KB]
   Test 2: TEST OK [0.000 secs, 3344 KB]
   Test 3: TEST OK [0.000 secs, 3344 KB]
   Test 4: TEST OK [0.000 secs, 3344 KB]
   Test 5: TEST OK [0.000 secs, 3344 KB]

All tests OK.

Your program ('sprime') produced all correct answers! This is your submission #2 for this problem. Congratulations!

Here are the test data inputs:

------- test 1 ----
4
------- test 2 ----
5
------- test 3 ----
6
------- test 4 ----
7
------- test 5 ----
8
Keep up the good work!


Thanks for your submission!

 

 

Super Prime RibRuss Cox

We use a recursive search to build superprimes of length n from superprimes of length n-1 by adding a 1, 3, 7, or 9. (Numbers ending in any other digit are divisible by 2 or 5.) Since there are so few numbers being tested, a simple primality test suffices.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

FILE *fout;

int
isprime(int n)
{
	int i;

	if(n == 2)
		return 1;

	if(n%2 == 0)
		return 0;

	for(i=3; i*i <= n; i+=2)
		if(n%i == 0)
			return 0;

	return 1;
}

/* print all sprimes possible by adding ndigit digits to the number n */
void
sprime(int n, int ndigit)
{
	if(ndigit == 0) {
		fprintf(fout, "%d\n", n);
		return;
	}

	n *= 10;
	if(isprime(n+1))
		sprime(n+1, ndigit-1);
	if(isprime(n+3))
		sprime(n+3, ndigit-1);
	if(isprime(n+7))
		sprime(n+7, ndigit-1);
	if(isprime(n+9))
		sprime(n+9, ndigit-1);
}

void
main(void)
{
	int n;
	FILE *fin;

	fin = fopen("sprime.in", "r");
	assert(fin != NULL);
	fout = fopen("sprime.out", "w");
	assert(fout != NULL);

	fscanf(fin, "%d", &n);

	sprime(2, n-1);
	sprime(3, n-1);
	sprime(5, n-1);
	sprime(7, n-1);
	exit (0);
}


 

### USACO 1.5 回文质数 Problem Solution #### 题目概述 给定一个整数范围,找出该范围内所有的既是回文又是质数的数字并输出。 #### 方法一:素数筛法结合回文判断 此方法先通过埃拉托斯特尼筛法预处理一定范围内的所有质数,再逐一验证这些质数是否为回文数[^1]。 ```cpp #include <iostream> #include <vector> using namespace std; bool isPalindrome(int n) { string str = to_string(n); int len = str.length(); for (int i = 0; i < len / 2; ++i) if (str[i] != str[len - 1 - i]) return false; return true; } const int MAXN = 1e6 + 5; vector<int> primes; void sieve() { vector<bool> prime(MAXN, true); for (long long p = 2; p * p < MAXN; ++p) if (prime[p]) for (long long multiple = p * p; multiple < MAXN; multiple += p) prime[multiple] = false; for (int p = 2; p < MAXN; ++p) if (prime[p] && isPalindrome(p)) primes.push_back(p); } ``` 上述代码实现了对指定区间内所有满足条件的数值进行筛选的功能。首先定义了一个辅助函数`isPalindrome()`用于检测某个正整数n是不是回文结构;接着利用布尔数组标记合数位置完成初步过滤工作,在此基础上进一步挑选出符合条件的目标对象加入到最终的结果列表当中去。 #### 方法二:直接构造特定长度的回文序列 考虑到题目特殊性质(即所求解必然是奇位数且回文),可以尝试按照固定模式构建候选集,之后仅需检验其可除性即可确认是否属于目标集合成员之一[^3]。 ```cpp for (int d1 = 1; d1 <= 9; d1 += 2) { // 奇数才可能是素数 for (int d2 = 0; d2 <= 9; ++d2) { for (int d3 = 0; d3 <= 9; ++d3) { int palindrome = 10000*d1 + 1000*d2 + 100*d3 + 10*d2 + d1; bool flag = true; for (int j = 2; j*j <= palindrome; ++j) if (palindrome % j == 0){ flag = false; break; } if(flag) cout << palindrome << endl; } } } ``` 这段程序片段展示了如何基于三位模板生成五位长的可能答案,并对其进行简单的因式分解测试来决定保留与否的操作逻辑。
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