zoukankan      html  css  js  c++  java
  • HDU——1058Humble Numbers(找规律)

    Humble Numbers

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 21853    Accepted Submission(s): 9553


    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.
     
     
    代码:
    #include<iostream>
    #include<set>
    using namespace std;
    typedef long long LL;
    LL er=1,san=1,wu=1,qi=1;
    LL minn(const LL &a,const LL &b,const LL &c,const LL &d)
    {
    	LL t=a;
    	if(b<t)
    		t=b;
    	if(c<t)
    		t=c;
    	if(d<t)
    		t=d;
    	//下面都要用if而非else-if 出现重复的数时才可以多个数均被++ 
    	if(t==a)
    		er++;
     	if(t==b)
    		san++;
     	if(t==c)
    		wu++;
     	if(t==d)
    		qi++;
    	return t;
    }
    int main(void)
    {
    	LL i;
    	LL list[5900]={1,1};
    	for (i=2; i<5900; i++)
    	{
    		list[i]=minn(2*list[er],3*list[san],5*list[wu],7*list[qi]);
    	}			
    	int n;
    	while (cin>>n&&n)
    	{
    		if(n%10==1&&n%100!=11)
    			printf("The %dst humble number is %lld.
    ",n,list[n]);		
    		else if(n%10==2&&n%100!=12)
    			printf("The %dnd humble number is %lld.
    ",n,list[n]);
    		else if(n%10==3&&n%100!=13)
    			printf("The %drd humble number is %lld.
    ",n,list[n]);
    		else
    			printf("The %dth humble number is %lld.
    ",n,list[n]);
    	}
    	return 0;
    }
  • 相关阅读:
    模块:标准库Shelve
    模块:标准库shutil
    模块:标准库sys
    关于html中的文本节点问题
    MVVM
    iOS 检测有没有安装其它应用 和ios9下要注意的地方
    iOS9对SDK的影响(iOS9适配必看)
    CircleLayout
    MapSearch 阅读随笔
    苹果官网 demo The Elements 阅读随笔
  • 原文地址:https://www.cnblogs.com/Blackops/p/5356442.html
Copyright © 2011-2022 走看看