M

M - Humble Numbers

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers. 

Write a program to find and print the nth element in this sequence 

 

Input

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n. 

 

Output

For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output. 

 

Sample Input

 1234111213212223100100058420 

 

Sample Output

 The 1st humble number is 1.The 2nd humble number is 2.The 3rd humble number is 3.The 4th humble number is 4.The 11th humble number is 12.The 12th humble number is 14.The 13th humble number is 15.The 21st humble number is 28.The 22nd humble number is 30.The 23rd humble number is 32.The 100th humble number is 450.The 1000th humble number is 385875.The 5842nd humble number is 2000000000.
这个题这种打表的方法也是让人无奈哦
#include <cstdio>#include <iostream>using namespace std;int main(){    int n;    int i,j;    int dp[21][200]; for(i=0;i<21;i++)  for(j=0;j<200;j++)   if(j==0)    dp[i][j]=1;   else    dp[i][j]=0; for(n=2;n<21;n++)   //线条总数  for(i=1;i<=n-1;i++)  //相互平行的线的条数   for(j=0;j<200;j++) //依次遍历每种交点情况    if(dp[n-i][j]==1)      //n-i条不平行的线如果有j个交点     dp[n][j+(n-i)*i]=1;//加上与平行线的交点共有j+(n-i)*i个交点 while(~scanf("%d",&n)){  cout<<"0";  for(j=1;j<=n*(n-1)/2;j++)   if(dp[n][j])    cout<<" "<<j;  cout<<endl; } return 0;}