SDAU 课程练习3 1004

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Problem D

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 2 Accepted Submission(s) : 1

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：
将所有的以  2 3 5 7  因子的数，排好序放在数组里。
思路：
n = 5842 。直接求肯定会超时（因为还要排序） 
那么只能靠脑洞来写了。
num[ i ] = min { 2*i，3*j，5*k，7*m }
借鉴的别人思路，毕竟我的脑洞没这么大...
感想：
还好啦，没有什么太多的感想！哼，很不开心，为什么不让我养小仓鼠，还说我幼稚，哼！
AC代码：
#include<iostream>#include<cmath>#include<string.h>#include<stdio.h>using namespace std;int main(){   // freopen("r.txt","r",stdin);    long long int arr[6666];    arr[1]=1;    int a,b,c,d;    a=b=c=d=1;    int i;    for(i=2;i<5844;i++)    {        arr[i]=min(arr[a]*2,min(arr[b]*3,min(arr[c]*5,arr[d]*7)));        //cout<<arr[i]<<endl;        if(arr[i]==arr[a]*2)            a++;        if(arr[i]==arr[b]*3)            b++;        if(arr[i]==arr[c]*5)            c++;        if(arr[i]==arr[d]*7)            d++;    }    //cout<<arr[5200]<<endl;    int n;    while(~scanf("%d",&n))    {        if(n==0) break;        cout<<"The "<<n;        if(n%10==1&&n%100!=11)            cout<<"st humble number is ";        else if(n%10==2&&n%100!=12)            cout<<"nd humble number is ";        else if(n%10==3&&n%100!=13)            cout<<"rd humble number is ";        else            cout<<"th humble number is ";        cout<<arr[n]<<"."<<endl;    }}

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