hdu 1085 Humble Numbers（打表）

题目：一个数的素数因子是只有有2,3,5,7，叫做humble number，前20个 humble number为 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27...

求第n个humble number；

分析：打表，开一个数组HumbleNumber[6000][5],HumbleNumber[i][0],存humble number，HumbleNumber[i][1,2,3,4]存HumbleNumber[i][0]*2,3 ,5,7的值，

HumbleNumber[i]是比Humblenumber[i-1][0]大的一个数，从HumbleNumber[1~i][1~4]都搜一遍 找出来了，复杂度O（N^2）

#include<iostream>
#include<cstdio>
using namespace std;
long long BumbleNumber[6000][5];
int prime[5]={0,2,3,5,7};
int main()
{
int N;
for(int i=1;i<=10;i++)
BumbleNumber[i][0]=i;
for(int i=1;i<=10;i++)
for(int j=1;j<=4;j++)
BumbleNumber[i][j]=BumbleNumber[i][0]*prime[j];

for(int i=11;i<=5842;i++)
{
int temp=BumbleNumber[i-1][0],ans=2000000001;
for(int j=1;j<=i-1;j++)
{
for(int k=1;k<=4;k++)
if(BumbleNumber[j][k]>temp && BumbleNumber[j][k]<ans)
ans=BumbleNumber[j][k];
}
BumbleNumber[i][0]=ans;
for(int k=1;k<=4;k++)
{
if(BumbleNumber[i][0]*prime[k]>2000000001)
BumbleNumber[i][k]=2000000001;
else
BumbleNumber[i][k]=BumbleNumber[i][0]*prime[k];
}
}
while(cin>>N)
{
if(N==0)
{ break; }
if(N%10==1&&N%100!=11)
{
cout<<"The "<<N<<"st humble number is "<<BumbleNumber
[0]<<"."<<endl;
continue;
}
if(N%10==2&&N%100!=12)
{
cout<<"The "<<N<<"nd humble number is "<<BumbleNumber
[0]<<"."<<endl;
continue;
}
if(N%10==3&&N%100!=13)
{
cout<<"The "<<N<<"rd humble number is "<<BumbleNumber
[0]<<"."<<endl;
continue;
}
cout<<"The "<<N<<"th humble number is "<<BumbleNumber
[0]<<"."<<endl;
}
}

问题分析：参考了网上的算法技巧，这种打表技巧，很巧妙，是递增的顺序，所以时间复杂度为O(N).......很巧妙！！！转载的别人的代码

#include <iostream>
#include <cstring>
using namespace std;
int min(int a,int b){
if(a<b) return a;
else return b;
}
int main(void){
long D[5843];
memset(D,0,sizeof(D));
D[1]=1;
int i,i1=1,i2=1,i3=1,i4=1;
int t1,t2,t3,t4;
for(i=2;i<5843;i++){
t1=D[i1]*2;
t2=D[i2]*3;
t3=D[i3]*5;
t4=D[i4]*7;
D[i]=min(min(t1,t2),min(t3,t4));
if(D[i]==t1) i1++;
if(D[i]==t2) i2++;
if(D[i]==t3) i3++;
if(D[i]==t4) i4++;
}
int n;
cin>>n;
while(n){
if(n%10==1&&n%100!=11){
cout<<"The "<<n<<"st humble number is "<<D
<<"."<<endl;
}
else if(n%10==2&&n%100!=12){
cout<<"The "<<n<<"nd humble number is "<<D
<<"."<<endl;
}
else if(n%10==3&&n%100!=13){
cout<<"The "<<n<<"rd humble number is "<<D
<<"."<<endl;
}else{
cout<<"The "<<n<<"th humble number is "<<D
<<"."<<endl;
}
cin>>n;
}
system("pause");
return 0;
}