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POJ 刷题系列：1840. Eqs
传送门：1840. Eqs
题意：
等式 a1x31 a2x32 a3x33 a4x34 a5x35=0a_1x_1^3 a_2x_2^3 a_3x_3^3 a_4x_4^3 a_5x_5^3=0 ，其中ai∈−50,50,i=1,2,3,4,5a_i in -50, 50, i = 1, 2, 3, 4, 5，xi∈−50,50,xi≠0x_i in -50, 50, x_i neq 0 ，求满足 等式(x1,x2,x3,x4,x5)(x_1, x_2, x_3, x_4, x_5)的解的个数。
思路：
折半搜索，把等式分成两半，左半部分两个变量循环枚举，并用hash记录每个值出现的次数。枚举右半部分是否在hash中出现过，统计个数即可。int会MLE，所以采用char，节省内存。
代码如下：
代码语言：javascript复制import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Map;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201801/P1840.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int maxn = 12500000;
void read() {
int a = ni();
int b = ni();
int c = ni();
int d = ni();
int e = ni();
char[] hash = new char[maxn * 2];
for (int i = -50; i <= 50; i) {
if (i == 0) continue;
for (int j = -50; j <= 50; j) {
if (j == 0) continue;
int sum = a * i * i * i b * j * j * j;
hash[maxn - sum] ;
}
}
int tot = 0;
for (int i = -50; i <= 50; i) {
if (i == 0) continue;
for (int j = -50; j <= 50; j) {
if (j == 0) continue;
for (int k = -50; k <= 50; k) {
if (k == 0) continue;
int sum = c * i * i * i d * j * j * j e * k * k * k;
if (sum <= maxn && sum >= -maxn) {
tot = hash[sum maxn];
}
}
}
}
out.println(tot);
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\oxygen_workspace\Algorithm");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
read();
out.flush();
if (!oj){
System.out.println("[" (System.currentTimeMillis() - s) "ms]");
}
}
public boolean more(){
return in.hasNext();
}
public int ni(){
return in.nextInt();
}
public long nl(){
return in.nextLong();
}
public double nd(){
return in.nextDouble();
}
public String ns(){
return in.nextString();
}
public char nc(){
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext(){
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
static class D{
public static void pp(int[][] board, int row, int col) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < row; i) {
for (int j = 0; j < col; j) {
sb.append(board[i][j] (j 1 == col ? "n" : " "));
}
}
System.out.println(sb.toString());
}
public static void pp(char[][] board, int row, int col) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < row; i) {
for (int j = 0; j < col; j) {
sb.append(board[i][j] (j 1 == col ? "n" : " "));
}
}
System.out.println(sb.toString());
}
}
static class ArrayUtils {
public static void fill(int[][] f, int value) {
for (int i = 0; i < f.length; i) {
Arrays.fill(f[i], value);
}
}
public static void fill(int[][][] f, int value) {
for (int i = 0; i < f.length; i) {
fill(f[i], value);
}
}
public static void fill(int[][][][] f, int value) {
for (int i = 0; i < f.length; i) {
fill(f[i], value);
}
}
}
static class Num{
public static <K> void inc(Map<K, Integer> mem, K k) {
if (!mem.containsKey(k)) mem.put(k, 0);
mem.put(k, mem.get(k) 1);
}
}
}
