HDU 1058 Humble Numbers

Posted 2021-06-04 jpphy0

Humble Numbers

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

分析

• 数与质因式一一对应，不同的数的质因式不同，不同的质因式得到不同的数
• 按照素数的线性筛的思想，得到因数为2、3、5、7的不大于20 0000 0000的所有数，然后排序

代码

// hdu 1058 Humble Numbers
#include<bits/stdc++.h>
using namespace std;
#define MAX_N 6000
#define MAX_T 2000000000
long long dp[MAX_N]={1,2,3,5,7}, dp2[MAX_N]={1,2,3,5,7};
long long pri[4]={2,3,5,7};
int main(){
    long long n = 1, p = 5;
    while(n <= 5842){
        for(long long i = 0; i < 4; i++){
            if(pri[i] > dp2[n] || pri[i]*dp[n] > MAX_T) break;
            dp[p] = pri[i]*dp[n], dp2[p] = pri[i], p++;
        }
        n++;
    }
    sort(dp+1, dp + p);
    while(scanf("%d", &n), n != 0){
		if(n%10==1&&n%100!=11){
            printf("The %dst humble number is ",n);
        }
        else if(n%10==2&&n%100!=12){
            printf("The %dnd humble number is ",n);
        }
        else if(n%10==3&&n%100!=13){
            printf("The %drd humble number is ",n);
        }
        else{
            printf("The %dth humble number is ",n);
        }
		printf("%d.\\n", dp[n-1]);
	}
    return 0;
}