hdu 1058

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题意：求素因子只有2 3 5 7的数<br>zsd:5842 各种打表<br>#include<iostream>

#include<cstring>

using
namespace std;

__int64
a[6000];

int main()

{

    int
n;

    memset(a,0,sizeof(a));

    __int64
c=3000000000;

        a[1]=1;a[2]=2;a[3]=3;a[4]=4;

        for(int
i=5;i<=5842;i++)

        {

            c=3000000000;

            for(int
j=1;j<i;j++)

            {

                if(a[j]*2>a[i-1]&&a[j]*2<c)

                    c=a[j]*2;

                else
if(a[j]*3>a[i-1]&&a[j]*3<c)

                    c=a[j]*3;

                else
if(a[j]*5>a[i-1]&&a[j]*5<c)

                    c=a[j]*5;

                else
if (a[j]*7>a[i-1]&&a[j]*7<c)

                    c=a[j]*7;

                else

                    continue;

            }

            a[i]=c;

        }

        while(cin>>n&&n)

        {

        if(n % 10 == 1 && n % 100 != 11) 

              printf("The %dst humble number is %lld.\n",n ,a[n]); 

          else
if(n % 10 == 2 && n % 100 != 12) 

              printf("The %dnd humble number is %lld.\n",n ,a[n]); 

          else
if(n % 10 == 3 && n % 100 != 13) 

              printf("The %drd humble number is %lld.\n",n ,a[n]); 

          else

              printf("The %dth humble number is %lld.\n",n ,a[n]); 

        }

        return
0;

}<br>方法2： 比较经典<br>#include<iostream><br>#include<cstdio><br>#include<cstring><br>#include<algorithm><br>using
namespace std;<br>long
long num[60000];<br>int
b[4]={2,3,5,7};<br>long
long  min(long
long a,long
long b,long
long c,long
long d)<br>{<br>    a=a>b?b:a;<br>    c=c>d?d:c;<br>    return
a>c?c:a;<br>}<br>int
main()<br>{<br>    num[1]=1;<br>    int
i;<br>    int
l1=1,l2=1,l3=1,l4=1;<br>    for(i=2;i<=5842;i++)<br>    {<br>        num[i]=min(num[l1]*2,num[l2]*3,num[l3]*5,num[l4]*7);<br>        if(num[i]==num[l1]*2) l1++;<br>        if(num[i]==num[l2]*3) l2++;<br>        if(num[i]==num[l3]*5) l3++;<br>        if(num[i]==num[l4]*7) l4++;<br>    }<br>    int
n;<br>    while(scanf("%d",&n),n)<br>    {<br>        if(n % 10 == 1 && n % 100 != 11)<br>              printf("The %dst humble number is %lld.\n",n ,num[n]);<br>          else
if(n % 10 == 2 && n % 100 != 12)<br>              printf("The %dnd humble number is %lld.\n",n ,num[n]);<br>          else
if(n % 10 == 3 && n % 100 != 13)<br>              printf("The %drd humble number is %lld.\n",n ,num[n]);<br>          else<br>              printf("The %dth humble number is %lld.\n",n ,num[n]);<br>    }<br>    return
0;<br>}