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