动态规划：HDU1058——Humble Numbers（丑数问题）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1058

参考博客（规律很巧妙）

https://blog.csdn.net/qianchangdiyin/article/details/50787676

任何一个数分解到最后都是由一些质数或者1相乘。

代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
long long num[5845];
long long min(long long a,long long b,long long c,long long d)
{
    int min1,min2;
    min1=a<b?a:b;
    min2=c<d?c:d;
    if(min1>=min2)
        return min2;
    else
        return min1;
}
int main()
{
    int n;
    int i=1;
    num[1]=1;
    int z2,z3,z5,z7;
    z2=z3=z5=z7=1;
    while(i<5842)
    {
        num[++i]=min(num[z2]*2,num[z3]*3,num[z5]*5,num[z7]*7);
        if(num[i]==num[z2]*2)
            z2++;
        if(num[i]==num[z3]*3)
            z3++;
        if(num[i]==num[z5]*5)
            z5++;
        if(num[i]==num[z7]*7)
            z7++;
    }
    while(~scanf("%d",&n)&&n)
    {
        if(n%10==1&&n%100!=11)
            printf("The %dst humble number is %lld.\n",n,num[n]);
        else if(n%10==2&&n%100!=12)
            printf("The %dnd humble number is %lld.\n",n,num[n]);
        else if(n%10==3&&n%100!=13)
            printf("The %drd humble number is %lld.\n",n,num[n]);
        else
            printf("The %dth humble number is %lld.\n",n,num[n]);
    }
    return 0;
}