POJ 刷题系列：1837. Balance
题意：
一个天枰上有C个钩子，把M个砝码挂在这些C个钩子上，问多少种平衡的挂法。
思路： 首先考虑单个砝码在天枰的位置，可以得到C个状态，对应的平衡度为:G[0] * C[i]，记录这些初始平衡度的个数。这样当放入第二个砝码时，可以遍历这些平衡度，从而得到一个新的平衡度。此时，不同位置，不同质量的砝码可能拥有相同平衡度，个数直接顺延给下一平衡度状态：当前平衡度 G[1]∗C[i] for i∈[−15,15]当前平衡度 G[1]∗C[i] for i∈[−15,15]当前平衡度 G[1] * C[i] space for space i in [-15, 15]
详见https://blog.csdn.net/lyy289065406/article/details/6648094/
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/201805/P2187.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
void read() {
int C = in.nextInt();
int G = in.nextInt();
int[] CS = new int[C];
int[] GS = new int[G];
for (int i = 0; i < C; i) CS[i] = in.nextInt();
for (int j = 0; j < G; j) GS[j] = in.nextInt();
int[][] dp = new int[21][15001];
dp[0][7500] = 1;
for (int i = 1; i <= G; i) { // 一个物品 每个 钩子 都放过一遍
for (int j = 0; j <= 15000; j) {
if (dp[i - 1][j] > 0) {
for (int k = 0; k < C; k) {
dp[i][j GS[i - 1] * CS[k]] = dp[i - 1][j];
}
}
}
}
System.out.println(dp[G][7500]);
}
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);
}
}
}
