hdu1058 Humble Numbers 解题报告

各种水 题目是简单使用动态规划 使用已有的动态规划进行推导

先上代码

1

2

3

5

7

1 2

  4

3

5

7

1 2 3

  4

  6

5

7

最后注意输出 即可

/*
  Name:
  Copyright:
  Author:yujiaao
  Date:
  Description:hdu
*/
#include<iostream>
#include<cstdio>
using namespace std;
int list[5845];
int f_min(int a,int b,int c,int d)
{
    int m=a<b?a:b;
    int n=c<d?c:d;
    return n<m?n:m;
}
int main()
{
//    freopen("C:\\Users\\yujiaao\\Desktop\\测试专用\\in.txt","r",stdin);
//    freopen("C:\\Users\\yujiaao\\Desktop\\测试专用\\out.txt","w",stdout);
    list[0]=1;
    int i2=0,i3=0,i5=0,i7=0;
    int i=0;
    int temp;
    int n;
    for(i=1;i<5843;i++)
    {
        list[i]=f_min(2*list[i2],3*list[i3],5*list[i5],7*list[i7]);
        if(list[i]%2==0)
        i2++;
        if(list[i]%3==0)
        i3++;
        if(list[i]%5==0)
        i5++;
        if(list[i]%7==0)
        i7++;        
    }
    while(scanf("%d",&n)!=EOF &&n!=0)
    {
        if(n%10==1&&n%100!=11)
        printf("The %dst humble number is %d.\n",n,list[n-1]);
        else if(n%10==2&&n%100!=12)
        printf("The %dnd humble number is %d.\n",n,list[n-1]);
        else if(n%10==3&&n%100!=13)
        printf("The %drd humble number is %d.\n",n,list[n-1]);
        else
        printf("The %dth humble number is %d.\n",n,list[n-1]);
    }
    system("pause");
    return 0;
}