【杭电1058】Humble Numbers-动态规划

Humble NumbersCrawling in process...

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

SubmitStatus

Description

Input

Output

Sample Input

Sample Output

Hint

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.

题意：找出丑数：将一个数拆开，因数只有2，3，5，7；第一个数为1；找出所有的丑数，单调递增数列。

题解：累乘

           1*2 ..             2*2..           3*2..

           1*3 ..             2*3..           3*3..

      1    1*5 ..         2   2*5 ..       3  3*5..      5..        7..

           1*7...             2*7 ..          3*7..

code:

#include<cstdio>#include<algorithm>using namespace std;long long f[5843];int main(){int n;f[1]=1;int g[4]={2,3,5,7};for(int i=2;i<=5843;i++){f[i]=2000000001;//赋予最大值for(int j=0;j<4;j++){for(int k=i-1;k>=1;k--){if(f[k]*g[j]<=f[i-1])//比前一个还小跳出break;if(f[k]*g[j]<f[i])//插入使单增f[i]=f[k]*g[j];}}}while(scanf("%d",&n) && n){        printf("The %d",n);        if(n%10==1 && n%100!=11)            printf("st ");        else if(n%10==2 && n%100!=12)            printf("nd ");        else if(n%10==3 && n%100!=13)            printf("rd ");        else            printf("th ");        printf("humble number is %I64d.\n",f[n]);    }return 0;}