hdu1058 && hdu3199

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22763    Accepted Submission(s): 9950

Problem 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

   
   
   
   

    
    
    
    1
2
3
4
11
12
13
21
22
23
100
1000
5842
0
   
   
   
   

 

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.
   
   
   
   
   
   
   
   


   
   
   
   

要求的数必定满足 n = 2^a * 3^b * 5^c * 7^d

求第k个满足条件的数，利用动归的思想，每次只取前面记录的最小值，依次类推

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int NI = 5843;
long long dp[NI];
int n, a, b, c, d;
long long Min(long long a, long long b, long long c, long long d) {
    a = min(a, b);
    a = min(a, c);
    a = min(a, d);
    return a;
}
int main() {
    a = b = c = d = 0;
    dp[0] = 1;
    for(int i = 1; i < NI; i++) {
        dp[i] = Min(dp[a]*2, dp[b]*3, dp[c]*5, dp[d]*7);
        if(dp[a]*2 == dp[i]) a++; //2的个数
        if(dp[b]*3 == dp[i]) b++; //3的个数
        if(dp[c]*5 == dp[i]) c++; //5的个数
        if(dp[d]*7 == dp[i]) d++; //7的个数
    }
    while(~scanf("%d", &n) && n) {
        if(n % 100 == 0 || n % 100 == 11 || n % 100 == 12 || n % 100 == 13) {
            printf("The %dth humble number is %d.\n", n, dp[n-1]);
        }
        else {
            if(n % 10 == 1) {
                printf("The %dst humble number is %d.\n", n, dp[n-1]);
            } 
            else if(n % 10 == 2) {
                printf("The %dnd humble number is %d.\n", n, dp[n-1]);
            }
            else if(n % 10 == 3) {
                printf("The %drd humble number is %d.\n", n, dp[n-1]);
            }
            else {
                printf("The %dth humble number is %d.\n", n, dp[n-1]);
            }
        }
    }
    return 0;
}

~~~~~~~~~~~~~~~~~~~~~~~华丽的分割线~~~~~~~~~~~~~~~~~~~~~~~~~

Hamming Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1205    Accepted Submission(s): 518

Problem Description

For each three prime numbers p1, p2 and p3, let's define Hamming sequence Hi(p1, p2, p3), i=1, ... as containing in increasing order all the natural numbers whose only prime divisors are p1, p2 or p3. 

For example, H(2, 3, 5) = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, ... 

So H5(2, 3, 5)=6.

 

Input

In the single line of input file there are space-separated integers p1 p2 p3 i.

 

Output

The output file must contain the single integer - Hi(p1, p2, p3). All numbers in input and output are less than 10^18.

 

Sample Input

   
   
   
   

    
    
    
    7 13 19 100
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    26590291
   
   
   
   
   
   
   
   


   
   
   
   

同上题思路是一样的，只是这题的素数是键盘输入得到，因此要先对三个素数从小到大排序

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int NI = 1e5+5;
long long p[3];
long long dp[NI];
int n, a, b, c;
long long Min(long long a, long long b, long long c) {
    a = min(a, b);
    a = min(a, c);
    return a;
}
int main() {
    while(~scanf("%I64d%I64d%I64d%d", &p[0], &p[1], &p[2], &n)) {
        a = b = c = 0;
        dp[0] = 1;
        for(int i = 1; i <= n; i++) {
            dp[i] = Min(dp[a]*p[0], dp[b]*p[1], dp[c]*p[2]);
            if(dp[a]*p[0] == dp[i]) a++;
            if(dp[b]*p[1] == dp[i]) b++;
            if(dp[c]*p[2] == dp[i]) c++;
        }
        printf("%I64d\n", dp[n]);
    }
    return 0;
}