poj1840

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涉及算法：简单的哈希表

题目大意：对于给定的a1，a2，a3，a4，a5求出a1*x1^3+a2*x2^3+a3*x3^3+a4*x4^3+a5*x5^3=0的解的个数

即求-(a1*x1^3+a2*x2^3)=a3*x3^3+a4*x4^3+a5*x5^3的解的个数

先枚举出左边的的所有数，存放到散列中，为了便于寻址采用hsah[leftNum]=leftNum出现的次数，这样时间复杂度由O(n^5)降低为O(n^2+n^3)

用short开数组，占内存小

import java.util.Scanner;public class Main_1840 {static int[] a=new int[6];static short[] hash=new short[25000001];public static void main(String[] args) {Scanner in=new Scanner(System.in);for(int i=1;i<=5;i++){a[i]=in.nextInt();}getHash();solve();System.out.println(ans);}static void getHash(){for(int x1=-50;x1<=50;x1++){if(x1==0) continue;for(int x2=-50;x2<=50;x2++){if(x2==0) continue;hash[getLeftResult(x1, x2)]++;}}}static int ans=0;static int rightResult=0;static void solve(){for(int x3=-50;x3<=50;x3++){if(x3==0) continue;for(int x4=-50;x4<=50;x4++){if(x4==0) continue;for(int x5=-50;x5<=50;x5++){if(x5==0) continue;rightResult=getRightResult(x3, x4, x5);if(rightResult>=0 && rightResult<=25000000){ans+=hash[rightResult];}}}}}static int result=0;static int getLeftResult(int x1,int x2){result=0;result=a[1]*x1*x1*x1+a[2]*x2*x2*x2;return -result+12500000;//0~25000000}static int getRightResult(int x3,int x4,int x5){result=0;result=a[3]*x3*x3*x3+a[4]*x4*x4*x4+a[5]*x5*x5*x5;return result+12500000;}}

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