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可以软打表

Posted by KatrineYang at 2016-09-09 12:57:16 on Problem 1338 and last updated at 2016-09-09 12:59:00
先苯地暴力计祘一遍最大能到多少(大概用10幾秒吧),就是我代妈里的preprocess(),结果是859963392,然后㞎这个范围内面的2,3,5的幂次列出来再排序即可
注意第二遍计祘的时猴,需要long long int,这是因为尿然THRES苯身并没有超int,但是THRES*5超了,判断终止条件时猴需要,否则就死循环了
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;

long long int uglyNum[1505];

void preprocess(){
	int cnt = 1;
	for(int i = 1; ; i++){
		int I = i;
		while(I%2==0){
			I/=2;
		}
		while(I%3==0){
			I/=3;
		}
		while(I%5==0){
			I/=5;
		}
		if(I==1){
			uglyNum[cnt] = i;
			cnt++;
		}
		//cout << cnt << endl;
		if(cnt > 1500) break;
	}
}

const int THRES = 859963392;

void preprocess1(){
	int cnt = 1;
	long long int p2 = 1;
	while(1){
		long long int p3 = p2;
		while(1){
			long long int p5 = p3;
			while(p5 <= THRES){
				//cout << p5 << endl;
				uglyNum[cnt] = p5;
				cnt++;
				p5 *= 5;
			}
			p3 *= 3;
			if(p3 > THRES) break;
		}
		p2 *= 2;
		if(p2 > THRES) break;
	}
	sort(uglyNum+1, uglyNum+1501);
}

int main() {
	preprocess1();
	while(1){
		int n;
		scanf("%d", &n);
		if(n == 0) break;
		printf("%I64d\n", uglyNum[n]);
	}
	return 0;
}

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