Humble Numbers(hdu1058)

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Humble Numbers

TimeLimit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K(Java/Others)
Total Submission(s): 8304 Accepted Submission(s):3618


Problem Description
A number whose only prime factors are 2,3,5 or 7 is called ahumble 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 humblenumbers.

Write a program to find and print the nth element in thissequence

Input
The input consists of one or more test cases. Each test caseconsists 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 humblenumber is number.". Depending on the value of n, the correct suffix"st", "nd", "rd", or "th" for the ordinal number nth has to be usedlike it is shown in the sample output.

Sample Input
1 2 3 4 1112 13 21 22 23 100 1000 5842 0

Sample Output
The 1sthumble number is 1. The 2nd humble number is 2. The 3rd humblenumber is 3. The 4th humble number is 4. The 11th humble number is12. The 12th humble number is 14. The 13th humble number is 15. The21st humble number is 28. The 22nd humble number is 30. The 23rdhumble number is 32. The 100th humble number is 450. The 1000thhumble number is 385875. The 5842nd humble number is2000000000.

Source
University of Ulm Local Contest 1996

#include <iostream>#include <stdio.h> #include <math.h>     #include <stdio.h>    #include <stdlib.h>    #include <string.h>    #include <algorithm>   using namespace std;const int N=5843;int n,i;__int64 hum[N];__int64 min(__int64 a,__int64 b){    return a<b?a:b;}int main(){    longi1,i2,i3,i4;   i1=i2=i3=i4=1;   hum[1]=1;   for(i=2;i<N;i++){       hum[i]=min(hum[i1]*2,min(hum[i2]*3,min(hum[i3]*5,hum[i4]*7)));       if(hum[i]==hum[i1]*2) i1++;       if(hum[i]==hum[i2]*3) i2++;       if(hum[i]==hum[i3]*5) i3++;       if(hum[i]==hum[i4]*7) i4++;   }//用已经生成的数去乘以因子然后去最小的数生成新的数   while(scanf("%d",&n)!=EOF&&n){        if(n==1&&n0!=11)           printf("The %dst humble number is %I64d.\n",n,hum[n]);        else if(n==2&&n0!=12)           printf("The %dnd humble number is %I64d.\n",n,hum[n]);        else if(n==3&&n0!=13)           printf("The %drd humble number is %I64d.\n",n,hum[n]);        else           printf("The %dth humble number is %I64d.\n",n,hum[n]);    }    return0;}



 

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