Superprime Rib（DFS + 剪枝）

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本文介绍了一个算法问题，即找出所有N位的超级素数，这些素数的特点是从N位到1位的所有截断子串都是素数。文章提供了一种使用深度优先搜索(DFS)并结合有效剪枝策略的解决方案。

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)

OUTPUT FORMAT

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

SAMPLE OUTPUT (file sprime.out)

题意：

给出N，表示N位数（1到8），输出所有满足N，N~N-1，N~N-2……N~1均为素数的数。比如2333，2，23，233，2333均为素数，故输出2333。

思路：

DFS，剪枝。每一位构成均为素数，故每一位除了第一位之外不可能出现偶数或者5，所以素数判断也可以排除判断被偶数整除的可能。

AC：

/*   
TASK:sprime   
LANG:C++   
ID:sum-g1   
*/ 
#include<stdio.h>
#include<string.h>
int n,sum=0;

int test(int num)
{
    for(int i=3;i<=num/2;i+=2)
        if(!(num%i)) return 0;
    return 1;
}

void dfs(int dig,int k)
{
    sum=sum*10+k;
    if(dig>n) return;
    if(!test(sum))
    {
        return;
    }
    if(dig==n)
    {
        printf("%d\n",sum);
        return;
    }
    for(int i=1;i<=9;i+=2)
    {
        if(i==5) continue;
	dfs(dig+1,i);
        sum/=10;
    }
}

int main()
{
    freopen("sprime.in","r",stdin);      
    freopen("sprime.out","w",stdout);     
    int i=2;
    scanf("%d",&n);
    while(i)
    {
        dfs(1,i);
        sum=0;
        if(i==7) i=0;
        if(i==5) i=7;
        if(i==3) i=5;
        if(i==2) i=3;
    }
    return 0;
}