一类计算问题小结poj&hoj Set Definition ，Humble Numbers ，Ugly Numbers 因子构造法

/*这类问题都是给一些只能由某些特定因子乘积组成的数，然后求第n个数的大小。
做法就是每次用所给的条件来构造因子，预处理出前N项的值，复杂度O(N)，查询复杂度O(1).*/
Set Definition.code
#include <stdio.h>
#include <iostream>
using namespace std;
int a[10000000];
int main()
{
a[1]=1;
int n1=1,n2=1;
for(int i=2;i<=10000000;i++)
{
a[i]=min(2*a[n1]+1,3*a[n2]+1);
if(a[i]==2*a[n1]+1) n1++;
if(a[i]==3*a[n2]+1) n2++;
}
int x;
while(scanf("%d",&x)==1)
{
printf("%d\n",a[x]);
}
return 0;
}
Humble Numbers.code
#include <stdio.h>
#include <iostream>
using namespace std;
int a[6000];
int main()
{
a[1]=1;
int n2=1,n3=1,n5=1,n7=1;
for(int i=2; i<=5842; i++)
{
a[i]=min(min(a[n2]*2,a[n3]*3),min(a[n5]*5,a[n7]*7));
if(a[i]==a[n2]*2) n2++;
if(a[i]==a[n3]*3) n3++;
if(a[i]==a[n5]*5) n5++;
if(a[i]==a[n7]*7) n7++;
}
int n;
while(scanf("%d",&n)==1&&n)
{
if(n%10==1)
{
if(n%100!=11) printf("The %dst humble number is %d.\n",n,a
);
else printf("The %dth humble number is %d.\n",n,a
);
}
else if(n%10==2)
{
if(n%100!=12) printf("The %dnd humble number is %d.\n",n,a
);
else printf("The %dth humble number is %d.\n",n,a
);
}
else if(n%10==3)
{
if(n%100!=13) printf("The %drd humble number is %d.\n",n,a
);
else printf("The %dth humble number is %d.\n",n,a
);
}
else printf("The %dth humble number is %d.\n",n,a
);
}
return 0;
}
Ugly Number.code
#include <iostream>
using namespace std;
int main()
{
int number[1505];
int judge[3];
int i,j;
int n;
i=0;
number[0]=1;
while(i<=1500)
{
for( j=0;j<=i,number[j]*2<=number[i];j++);

judge[0]=number[j]*2;

for( j=0;j<=i,number[j]*3<=number[i];j++);

judge[1]=number[j]*3;

for( j=0;j<=i,number[j]*5<=number[i];j++);

judge[2]=number[j]*5;

number[i+1]=min(judge[0],min(judge[1],judge[2]));
i++;
}

cout <<"The 1500'th ugly number is " << number[1499] << "." << endl;

return 0;
}