HDU 1058 Humble Numbers-优快云博客

本文提供了一种求解第N个谦数的算法实现，使用C++编程语言，通过不断生成新的谦数并更新其最小值来找到指定位置的谦数。该算法适用于解决类似于HDU ACM 1058的问题。

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

#include<iostream>
#include<cstdio>
using namespace std;

int humble[6000];
int a1,a2,a3,a4;
int e1,e2,e3,e4;

int _min(int a,int b,int c,int d)
{
    int min;
    if(a<b)
        min=a;
    else
        min=b;
    if(min>c)
        min=c;
    if(min>d)
        min=d;
    return min;
}

int main()
{
    int n,i;
    a1=a2=a3=a4=1;
    humble[1]=1;
    for(i=2;i<6000;i++)
    {
        e1=humble[a1]*2;
        e2=humble[a2]*3;
        e3=humble[a3]*5;
        e4=humble[a4]*7;
        humble[i]=_min(e1,e2,e3,e4);
        if(humble[i] == e1)
            a1++;
        if(humble[i] == e2)
            a2++;
        if(humble[i] == e3)
            a3++;
        if(humble[i] == e4)
            a4++;
    }

     while(scanf("%d",&n)&& n)
    {
        if(n%100==11)
        printf("The %dth humble number is %d.\n",n,humble[n]);
        else if(n%100==12)
        printf("The %dth humble number is %d.\n",n,humble[n]);
        else if(n%100==13)
        printf("The %dth humble number is %d.\n",n,humble[n]);
        else if(n%10==1)
        printf("The %dst humble number is %d.\n",n,humble[n]);
        else if(n%10==2)
        printf("The %dnd humble number is %d.\n",n,humble[n]);
        else if(n%10==3)
        printf("The %drd humble number is %d.\n",n,humble[n]);
        else
        printf("The %dth humble number is %d.\n",n,humble[n]);
    }
 return 0;
}