acm_Humble Numbers

最新推荐文章于 2019-03-27 22:07:28 发布

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本文介绍了一种算法，用于计算特定数列中的Humble Numbers，这些数的质因数仅包含2、3、5和7，并提供了一个高效的C++实现方案。

题目：

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

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：每输入一个数n，让求从1开始的第n个Humble Numbers是多少，而Humble Numbers是因子只有2 3 5 7这几个数，当然1也算。。

思路：可以先用一个vector容器把这5842个数求出来。。然每输入一个数取就可以了。。

代码：

#include <iostream>
#include <string>
#include<stdio.h>
#include<vector>
#include<cmath>
using namespace std;
int inline min(int a,int b)
{
return a<b?a:b;
}
int main(){
int n = 5842, m;
char s[4][3]={"th","st","nd","rd"};
vector<int> v;
int a = 0, b = 0, c = 0, d = 0;
v.push_back(1);
while(--n){
int r2 = v[a] * 2;
int r3 = v[b] * 3;
int r5 = v[c] * 5;
int r7 = v[d] * 7;
m = min(r2, r3);
m = min(m, r5);
m = min(m, r7);
v.push_back(m);
if(r2 == m)
a+=1;
if(r3 == m)
b+=1;
if(r5 == m)
c+=1;
if(r7 == m)
d+=1;
}
while(cin>>n&&n!=0)
{
string cc=(n % 10 < 4 && n % 100 != 11 && n % 100 != 12 && n % 100 != 13)?s[n % 10]:"th";
cout<<"The "<<n<<cc<<" humble number is "<<v[n - 1]<<"."<<endl;
}
return 0;
}