POJ 1338 解题报告

来源：互联网发布：sql数据库卸载编辑：程序博客网时间：2024/05/29 13:08

这道题我是自己写的二分查找加Java的BigInteger做的。看了后面打表的做法，大整数似乎是不需要的，int就足够了。

实际上二分查找维护一个排序的数组就是priority queue的主要作用，用STL可以大大简化过程。

基于以上两点，时间很慢。如果思路不变，但用STL+int，discuss上面有0ms的。

最聪明的做法应该是这样的：

http://poj.org/showmessage?message_id=113667

维护3个指针齐头并进，i2_mul指向可以乘2的最小的数（换句话说比这个数大的数就没必要试试乘2了）。其他两个变量同理。谁选上了（最小），谁更新下（++）。

thestoryofsnow1338Accepted6280K282MSJava2004B

import java.io.*;import java.util.*;import java.math.BigInteger;public class Poj1338 {public static void main(String args[]){final int MAXN = 1500;final BigInteger TWO = BigInteger.valueOf(2);final BigInteger THREE = BigInteger.valueOf(3);final BigInteger FIVE = BigInteger.valueOf(5);final BigInteger[] BASE = new BigInteger[3];BASE[0] = TWO; BASE[1] = THREE; BASE[2] = FIVE;BigInteger[][] powers = new BigInteger[3][MAXN];for (int i = 0; i < 3; ++i){powers[i][0] = BigInteger.ONE;for (int n = 1; n < MAXN; ++n){powers[i][n] = powers[i][n - 1].multiply(BASE[i]);// if (n < 5)// {// System.out.println("powers[" + i + "][" + n + "]: " + powers[i][n]);// }}}BigInteger[] nums = new BigInteger[MAXN];int nnums = 0;int low, high, mid;for (int x = 0; x < MAXN; ++x){for (int y = 0; (1 + x) * (1 + y) <= MAXN; ++y){for (int z = 0; (1 + x) * (1 + y) * (1 + z) <= MAXN; ++z){BigInteger num = powers[0][x].multiply(powers[1][y]).multiply(powers[2][z]);low = 0; high = nnums - 1;while (low <= high){mid = low + (high - low) / 2;if (nums[mid].compareTo(num) <= 0){low = mid + 1;}else{high = mid - 1;}}if (low == MAXN){continue;}else{if (nnums < MAXN){nnums++;}for (int i = nnums - 1; i >= low + 1; --i){nums[i] = nums[i - 1];}nums[low] = num;}}}}// Collections.sort(nums);// for (int i = 0; i < 10; ++i)// {// System.out.print(nums[i] + " ");// }// System.out.println();Scanner cin = new Scanner(System.in);Integer n;while (cin.hasNext()){n = cin.nextInt();if (n == 0){break;}System.out.println(nums[n - 1]);}}}

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