hdu1058 Humble Numbers DP

最新推荐文章于 2019-09-20 00:05:18 发布

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本文介绍了一种高效算法来寻找Humble Numbers，即那些仅由2、3、5、7作为质因数的数。通过动态规划的方法，该算法能够快速定位到序列中的任意一项，并给出完整的C++实现代码。

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26228 Accepted Submission(s): 11560

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

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.

题目大意就是，2,3,5,7，被称为humble number，因子只含2,3,5,7的也被称为humble number，求第n个humble number数

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=5843;
int dp[N];

void fun(){
	dp[1]=1;
	int i=1,j=1,k=1,p=1,q=2;
	int a,b,c,d;
	for(;q<5843;q++){
		dp[q] = min(min(a=dp[i]*2,b=dp[j]*3),min(c=dp[k]*5,d=dp[p]*7));
		if(a==dp[q])i++;
		if(b==dp[q])j++;
		if(c==dp[q])k++;
		if(d==dp[q])p++;
	}
}

int main(){
	int n;
	fun();
	while(~scanf("%d",&n)&&n){
		if(n%100!=11 && n%10==1){
			printf("The %dst humble number is %lld.\n",n,dp[n]);
		}
		else if(n%100!=12 && n%10==2){
			printf("The %dnd humble number is %lld.\n",n,dp[n]);
		}
		else if(n%100!=13 && n%10==3){
			printf("The %drd humble number is %lld.\n",n,dp[n]);
		}
		else{
			printf("The %dth humble number is %lld.\n",n,dp[n]);
		}
	}
	return 0;
}