hdu 1058（简单dp）Humble Numbers

最新推荐文章于 2018-04-16 16:05:00 发布

forwiat

最新推荐文章于 2018-04-16 16:05:00 发布

阅读量310

点赞数

CC 4.0 BY-SA版权

分类专栏： dp 文章标签： hdu1058简单dp Humble Numbers

本文链接：https://blog.youkuaiyun.com/mystery_guest/article/details/52849510

dp 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一种特殊的数——谦逊数，这些数仅由2、3、5、7这几个质因数构成。通过动态规划算法，可以高效地找出序列中的任意一项。文中详细解释了算法的实现，并给出了一段C++代码示例。

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24184 Accepted Submission(s): 10590

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.

简单dp搞一搞，在外面打个表即可。
dp思想：从第一位开始往5842推，第i位肯定是前面i-1位乘以2，3，5，7四个数中一个的最小数，所以，用4个数来分别标记最后乘以2，3，5，7的数。
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<stack>
#include<queue>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#define eps 1e-8
#define zero(x) (((x>0?(x):-(x))-eps)
#define mem(a,b) memset(a,b,sizeof(a))
#define memmax(a) memset(a,0x3f,sizeof(a))
#define pfn printf("\n")
#define ll __int64
#define ull unsigned long long
#define mod 1000000007
#define sf(a) scanf("%d",&a)
#define sf64(a) scanf("%I64d",&a)
#define sf264(a,b) scanf("%I64d%I64d",&a,&b)
#define sf364(a,b,c) scanf("%I64d%I64d%I64d",&a,&b,&c)
#define sf2(a,b) scanf("%d%d",&a,&b)
#define sf3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define sf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)
#define sff(a) scanf("%f",&a)
#define sfs(a) scanf("%s",a)
#define sfs2(a,b) scanf("%s%s",a,b)
#define sfs3(a,b,c) scanf("%s%s%s",a,b,c)
#define sfd(a) scanf("%lf",&a)
#define sfd2(a,b) scanf("%lf%lf",&a,&b)
#define sfd3(a,b,c) scanf("%lf%lf%lf",&a,&b,&c)
#define sfd4(a,b,c,d) scanf("%lf%lf%lf%lf",&a,&b,&c,&d)
#define sfc(a) scanf("%c",&a)
#define ull unsigned long long
#define debug printf("***\n")
const double PI = acos(-1.0);
const double e = exp(1.0);
const int INF = 0x7fffffff;;
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> inline T Min(T a, T b) { return a < b ? a : b; }
template<class T> inline T Max(T a, T b) { return a > b ? a : b; }
bool cmpbig(int a, int b){ return a>b; }
bool cmpsmall(int a, int b){ return a<b; }
using namespace std;
#define MAX 100010
int main()
{
  //  freopen("data.in","r",stdin);
    int n,i;
    int hnum[6000];
    hnum[1]=1;
    int a=1,b=1,c=1,d=1;
    for(i=2;i<=5842;i++)
    {
        hnum[i]=min(hnum[a]*2,min(hnum[b]*3,min(hnum[c]*5,hnum[d]*7)));
        if(hnum[i]==hnum[a]*2) a++;
        if(hnum[i]==hnum[b]*3) b++;
        if(hnum[i]==hnum[c]*5) c++;
        if(hnum[i]==hnum[d]*7) d++;
    }
    while(~sf(n))
    {
        if(n==0)
            break;
        printf("The %d",n);
        if(n%10==1&&n%100!=11)
            printf("st");
        else if(n%10==2&&n%100!=12)
            printf("nd");
        else if(n%10==3&&n%100!=13)
            printf("rd");
        else
            printf("th");
        printf(" humble number is %d.\n",hnum[n]);
    }
    return 0;
}