普通的方法:
package com.wangzhu.njupt; import java.io.BufferedInputStream; import java.io.IOException; import java.io.StreamTokenizer; /** * 由于输入的数在1~100000之间,故只需要开个100000大小的数组,判断其是否存在即可 * * @ClassName: Main1402 * @Description: TODO * @author 王竹 * @date 2014-10-3 下午5:21:31 * */ public class Main1402 { private static final int LEN = 1000001; /** * @param args * @throws IOException */ public static void main(String[] args) throws IOException { // System.setIn(new FileInputStream("data.in")); // 输入优化 StreamTokenizer in = new StreamTokenizer(new BufferedInputStream( System.in)); // Scanner cin = new Scanner(new BufferedInputStream(System.in)); // while (cin.hasNext()) { while (in.nextToken() != StreamTokenizer.TT_EOF) { int[] arr = new int[LEN]; // int n = cin.nextInt(); int n = (int) in.nval; int m = -1; for (int i = 0; i < n; i++) { // m = cin.nextInt(); in.nextToken(); m = (int) in.nval; arr[m]++; } StringBuilder sb = new StringBuilder(); int c = 0, index = 0; for (int i = 1; i < LEN; i++) { if (arr[i] == 1) { if (index == 0) { sb.append(i); index = 1; } else { sb.append(" ").append(i); } c++; } } System.out.println(c); if (c > 0) { System.out.println(sb); } } } }
位存储的方法:
package com.wangzhu.njupt; import java.io.BufferedInputStream; import java.io.FileInputStream; import java.io.IOException; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.io.StreamTokenizer; import java.util.Scanner; public class Main14022 { /** * 用位来存储数 */ private static final int LEN = (1000000 >> 3) + 10; /** * 用于模8,即与7 */ private static final int MASK = (1 << 3) - 1; /** * 要查的数的范围为1~1000000 */ private static final int SIZE = 1000001; /** * @param args * @throws IOException */ public static void main(String[] args) throws IOException { // System.setIn(new FileInputStream("data.in")); // Scanner cin = new Scanner(new BufferedInputStream(System.in)); // 输入优化 StreamTokenizer in = new StreamTokenizer(new BufferedInputStream( System.in)); // 输出优化 PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); // while (cin.hasNext()) { while (in.nextToken() != StreamTokenizer.TT_EOF) { int[] arr = new int[LEN]; int[] flag = new int[LEN]; // int n = cin.nextInt(); int n = (int) in.nval; for (int i = 0; i < n; i++) { // int m = cin.nextInt(); in.nextToken(); int m = (int) in.nval; // 其实是除以8 int index = m >> 3; // 其实是模8 int bit = m & MASK; if ((arr[index] & (1 << bit)) == 0) { // 记录输入中的数 arr[index] |= (1 << bit); } else if ((flag[index] & (1 << bit)) == 0) { // 记录输入多个同样的数 flag[index] |= (1 << bit); } } int cnt = 0; for (int i = 1; i < SIZE; i++) { int index = i >> 3; int bit = i & MASK; if ((arr[index] & (1 << bit)) != 0 && (flag[index] & (1 << bit)) == 0) { // 判断是输入的数并且只输入了一次 // 记录这种数的个数 cnt++; } } StringBuilder sb = new StringBuilder(); int j = 0; for (int i = 1; (i < SIZE) && (j < cnt); i++) { int index = i >> 3; int bit = i & MASK; if ((arr[index] & (1 << bit)) != 0 && (flag[index] & (1 << bit)) == 0) { sb.append(i).append(" "); j++; } } String ret = null; out.println(cnt); if (cnt > 0) { ret = sb.substring(0, sb.length() - 1); out.println(ret); } out.flush(); } } }
备注:
关于位运算的一些补充:
例如,
关于n/8,可以转化为n>>3。
关于n%8,可以转化为n&7,当然也已转化为n - (n>> 3)<<3。