HDU 1164 Eddy's research I（素数）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1164

题意：将n分解为素因子的形势。

思路：

一：将素数打表再暴力求解。

代码如下：

#include <iostream>using namespace std;#include <string.h>#include <stdio.h>#include <climits>#include <algorithm>#define maxn  65535typedef __int64 LL;int a[maxn];int prime[maxn];int main(){    memset(a,1,sizeof(a));    for( int i = 2; i <= maxn; i ++ )    {        if( a[i] )        for( int j = i+i; j <= maxn; j += i )             a[j]=0;    }    int j = 0;    for( int i = 2; i < maxn; i ++ )    {       if( a[i] ) prime[j] = i,j++;    }    int n;    while( scanf ( "%d", &n ) != EOF )    {        int f = 1;        for( int i = 0; i < j; i ++ )        {            if( n == 1 ) break;            while( n%prime[i] == 0 )            {                if( f )                {                    printf("%d",prime[i]);                    f=0;                }                else printf("*%d",prime[i]);                n /= prime[i];            }        }        printf("\n");    }    return 0;}

二：打表后用DFS遍历求解。

代码如下：

#include <iostream>using namespace std;#include <string.h>#include <stdio.h>#include <climits>#include <algorithm>#define maxn  65535typedef __int64 LL;int a[maxn], n, k;int prime[maxn];void DFS( int x ){    if( x == 0 ) return ;    for( int i = 0; i < k; i ++ )    {        if( x%prime[i] == 0 )        {            printf( x==n?"%d":"*%d",prime[i] );            DFS( x/prime[i] );            break;        }    }}int main(){    memset(a,1,sizeof(a));    for( int i = 2; i <= maxn/2; i ++ )    {        if( a[i] )        for( int j = i+i; j <= maxn; j += i )             a[j]=0;    }    k = 0;    for( int i = 2; i < maxn; i ++ )    {       if( a[i] ) prime[k++] = i;    }    while( scanf ( "%d", &n ) != EOF )    {        DFS( n );        printf("\n");    }    return 0;}

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