DP 60题 -3 HDU1058 Humble Numbers DP求状态数的老祖宗题目

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 32718    Accepted Submission(s): 14308

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.

Source

University of Ulm Local Contest 1996

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不给题意，直接贴代码，得好好看看！

#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
//---------------------------------Sexy operation--------------------------//

#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define debug(n) printf("%d_________________________________\n",n);
#define speed ios_base::sync_with_stdio(0)
#define file  freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
//-------------------------------Actual option------------------------------//
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Swap(a,b) a^=b^=a^=b
#define Max(a,b) (a>b?a:b)
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define mp(a,b) make_pair(a,b)
#define pb(n)  push_back(n)
#define dis(a,b,c,d) ((double)sqrt((a-c)*(a-c)+(b-d)*(b-d)))
//--------------------------------constant----------------------------------//

#define INF  0x3f3f3f3f
#define esp  1e-9
#define PI   acos(-1)
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef  long long ll;
//___________________________Dividing Line__________________________________/
long  long dp[6000];
long long a[4]={2,3,5,7};
int main()
{
    dp[1]=1;
    int p[4]={1,1,1,1};
    for(int i=2;i<=6000;i++)
    {
        int flag;
        long long ma=(1LL)<<60;
        for(int j=0;j<4;j++)
        {
            if(dp[p[j]]*a[j]<ma){ flag=j;
            ma=dp[p[j]]*a[j];}
        }
        p[flag]++;
        dp[i]=ma;
        if(dp[i]==dp[i-1])i--;
        //cout<<dp[i]<<endl;
    }
    int n;
    string s;
    while(cin>>n&&n)
    {

            if(n%10==1&&n%100!=11)
            s="st";
            else if(n%10==2&&n%100!=12)
            s="nd";
            else if(n%10==3&&n%100!=13)
            s="rd";
            else s="th";

        cout<<"The "<<n<<s<<" humble number is "<<dp[n]<<"."<<endl;
    }
    return 0;
}