USACO 1.5.3 sprime ———— 暴枚 || DFS

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

我们可以发现一个规律就是n等于1时只能是2,3,5,7,n>1时每一位都是1,3,7,9之一,那么我们直接暴枚即可,这题正解应该是DFS去搜每一位的数字。

/*
     ID: xinming2
     PROG: sprime
     LANG: C++
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,n) for(i=0;i<(n);++i)
#define FOR(i,l,h) for(i=(l);i<=(h);++i)
#define FORD(i,h,l) for(i=(h);i>=(l);--i)

typedef vector VI;
typedef vector VS;
typedef vector VD;
typedef long long LL;
typedef pair PII;
int q[10000] = {2 , 3 , 5,  7};
bool isprime(int x)
{
    if(x == 0 || x == 1)return 0;
    else for(int i = 3 ; i * i <= x ; i += 2)
    {
        if(x % i == 0)return 0;
    }
    return 1;
}
int main()
{
    freopen("sprime.in" , "r" , stdin);
    freopen("sprime.out" , "w" , stdout);
    int n;
    while(~scanf("%d", &n))
    {
        int maxnum = pow(10.0 , n);
        int minnum = maxnum / 10;
        int front = 0 , rear = 4;
        while(front < rear)
        {
            int k = q[front++];
            if(isprime(k * 10 + 1))q[rear++] = k * 10 + 1;
            if(isprime(k * 10 + 3))q[rear++] = k * 10 + 3;
            if(isprime(k * 10 + 7))q[rear++] = k * 10 + 7;
            if(isprime(k * 10 + 9))q[rear++] = k * 10 + 9;
            if(q[rear - 1] > maxnum)break;
        }
        for(int i = 0 ; q[i] < maxnum ; i++)
        {
            if(q[i] >= minnum)cout << q[i] << endl;
        }
    }
    return 0;
}


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