【CUGBACM15级BC第19场 A】hdu 5108 Alexandra and Prime Numbers

Alexandra and Prime Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2291    Accepted Submission(s): 755

Problem Description

Alexandra has a little brother. He is new to programming. One day he is solving the following problem: Given an positive integer N, judge whether N is prime.
The problem above is quite easy, so Alexandra gave him a new task: Given a positive integer N, find the minimal positive integer M, such that N/M is prime. If such M doesn't exist, output 0.
Help him!

 

Input

There are multiple test cases (no more than 1,000). Each case contains only one positive integer N.
N≤1,000,000,000.
Number of cases with N>1,000,000 is no more than 100.

 

Output

For each case, output the requested M, or output 0 if no solution exists.

 

Sample Input

3456

 

Sample Output

1212

 

题意：
     给一个n(<=10Y)，然后让找到一个最小的m使得n/m是一个素数.

思路：
      先用sqrt(n)的时间把所有的因子都求出来，然后在排序，枚举，就行了

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#include <ctime>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;const int NP = 32000;int ispri[NP] = {}, prime[NP], pcnt = 0;void getprime(){    ispri[0] = ispri[1] = 1;    for (long long i = 2; i < NP; i++)        if (ispri[i] == 0)        {            prime[++pcnt] = i;            for (long long j = i * i; j < NP; j += i)            {                ispri[j] = 1;            }        }}int a[1000000] = {}, icnt = 0;void pdec(int n){    for (int i = 1; prime[i]*prime[i] <= n; i++)        if (n % prime[i] == 0)        {            a[++icnt] = prime[i];            n /= prime[i];            i--;        }    if (n != 1)    {        a[++icnt] = n;    }}int YZ[100000] , yzs;void DB(int now){    yzs = 0;    int maxx = (int)sqrt(now);    for (int i = 1 ; i <= maxx ; i ++)    {        if (now % i == 0)        {            YZ[++yzs] = i;            YZ[++yzs] = now / i;        }    }    if (maxx * maxx == now)    {        yzs --;    }}int dd(int x){    if (x <= 1)    {        return 0;    }    int i, m = floor(sqrt(x) + 0.5);    for (i = 2; i <= m; i++)        if (x % i == 0)        {            return 0;        }    return 1;}int main(){    int n;    while (cin >> n)    {        if (n <= 1)        {            printf("0\n");            continue;        }        int flag = 0;        DB(n);        sort(YZ + 1 , YZ + yzs + 1);        for (int i = 1; i <= yzs; i++)        {            int xx = n / YZ[i];            if (dd(xx))            {                cout << YZ[i] << endl;                flag = 1;                break;            }        }        if (flag == 0)        {            cout << "0" << endl;        }    }    return 0;}

我们还是用另外一种方法去处理这道题，那就是质因数分解。

质因数分解：就是把一个合数分解成几个素数相乘的形式。例如：
48 = 2*2*2*3

58 = 2*3*3*3

由此我们可以将N一直除以一个比它自己本身小的质数，按照质因数分解的方法，不停的进行分解，我们要找到一个分解出的最大的素数P,因为只有这样，我们才能使得M最小。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <stack>#include <queue>#include <string>#include <vector>#include <set>#include <map>#include <bitset>#include <cassert>using namespace std;typedef long long LL;const int N = 50;int n;void work(){    int i , m , x = 0;    m = n;    for (i = 2 ; i * i <= n ; ++ i)    {        if (n % i == 0)        {            while (n % i == 0)            {                n /= i;            }            x = max(x , i);        }    }    if (n > 1)    {        x = max(x , n);    }    printf("%d\n" , x ? m / x : 0);}int main(){    while (~scanf("%d", &n))    {        work();    }    return 0;}