目录
Divide by Three：
题目大意：
编辑编辑思路解析：
代码实现：
Divide by Three：
题目大意：
思路解析：
一个数字是3的倍数，那么他的数位之和也是3的倍数，所以我们可以O(N)的得到这个数字数位之和模3的结果，如果模3为0，则这个数就是一个美丽的数，如果模3不为0，则他需要取除某些数字.
那我们想它最多去除几个数字，最多取除两个，因为在模3系统下，任意1-9的数都会等价与1-2，当总体剩下1时，只要有1就删除1，否则就删除两个2，当总体剩下2时，只要有2就删除2，否则就删除两个1，这里删除的1和2是等价的1和2.
删除之后我们比较这两个情况那个更优秀，删除时从后往前删，因为如果从前往后删，我们还需要考虑删除这个数，后面的是否会变成前缀为0，但是如果从后往前删，就不需要考虑，除非它只能删除这个数使得前缀为0。
还有特殊情况，删除后没数字剩余了，或者删除后只剩下0了；
代码实现：
import java.beans.IntrospectionException;
import java.io.*;
import java.math.BigInteger;
import java.util.*;public class Main {static int tot = 0;static int maxn = 123460;static int[][] ch;static int[] tag;static long ans = 0;static long[] sum = new long[maxn];static int[] arr = new int[maxn];static HashMap<Integer, ArrayList<Integer>> map;static int k;public static void main(String[] args) throws IOException {char[] s = f.next().toCharArray();int n = s.length;int sum = 0;int[] cnt = new int[3];for (int i = 0; i < n; i++) {sum = (sum + s[i] - '0') % 3;cnt[(s[i] - '0') % 3]++;}if (sum == 0){for (int i = 0; i < n; i++) {w.print(s[i]);}w.flush();w.close();return;}char[] ans = new char[n];char[] res = new char[n];int p = 0;int q = 0;if (sum == 1){if (cnt[1] >= 1){int tag = 1;for (int i = n-1; i >= 0; i--) {if ((s[i] - '0') % 3 == 1 && tag >= 1){tag--;}else ans[p++]= s[i];}}if (cnt[2] >= 2){int tag = 2;for (int i = n-1; i >= 0; i--) {if ((s[i] - '0') % 3 == 2 && tag >= 1){tag--;}else res[q++]= s[i];}}}else {if (cnt[1] >= 2){int tag = 2;for (int i = n-1; i >= 0; i--) {if ((s[i] - '0') % 3 == 1 && tag >= 1){tag--;}else ans[p++]= s[i];}}if (cnt[2] >= 1){int tag = 1;for (int i = n-1; i >= 0; i--) {if ((s[i] - '0') % 3 == 2 && tag >= 1){tag--;}else res[q++]= s[i];}}}int flag1 = 0, flag2 = 0;while (p >= 1 && ans[p - 1] == '0') {p--; flag1 = 1;}while (q >= 1 && res[q - 1] == '0') {q--; flag2 = 1;}if (flag1 == 1 && p == 0) ans[p++] ='0';if (flag2 == 1 && q == 0) res[q++] ='0';if (p == 0 && q == 0) w.println(-1);else {if (q > p){for (int i = q - 1; i >= 0; i--) {w.print(res[i]);}}else {for (int i = p - 1; i >= 0; i--) {w.print(ans[i]);}}}w.flush();w.close();}public static void get(int x, int r, int p , int cur, long res){if (res >= k && tag[p] == 1){ArrayList<Integer> list = map.get(p);for (Integer l : list) {ans = Math.max(ans, (sum[r] - sum[l - 1]) / (r - l + 1));}}if (cur < 0)return;int c = (x >> cur) & 1;if (ch[p][1 - c] != 0 && res + (1L << (cur + 1)) - 1>= k) get(x, r, ch[p][1 - c], cur-1, res + (1L <<cur));if (ch[p][c] != 0 && res + (1L << (cur)) - 1 >= k) get(x, r, ch[p][c], cur-1, res);}public static void insert(int x, int id){int p = 0;for (int i = 30; i >= 0; i--) {int c = ((x >> i) & 1);if (ch[p][c] == 0) ch[p][c] = ++ tot;p = ch[p][c];}tag[p] = 1;if (map.containsKey(p)){ArrayList<Integer> list = map.get(p);list.add(id);}else {ArrayList<Integer> list = new ArrayList<>();list.add(id);map.put(p, list);}}public static long qkm(long a, long mod){long res = 1;long b = mod - 2;while(b > 0){if ((b & 1) == 1) res = (res * a) % mod;a = (a * a) % mod;b >>= 1;}return res;}static PrintWriter w = new PrintWriter(new OutputStreamWriter(System.out));static Input f = new Input(System.in);static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));static class Input {public BufferedReader reader;public StringTokenizer tokenizer;public Input(InputStream stream) {reader = new BufferedReader(new InputStreamReader(stream), 32768);tokenizer = null;}public String next() {while (tokenizer == null || !tokenizer.hasMoreTokens()) {try {tokenizer = new StringTokenizer(reader.readLine());} catch (IOException e) {throw new RuntimeException(e);}}return tokenizer.nextToken();}public String nextLine() {String str = null;try {str = reader.readLine();} catch (IOException e) {// TODO 自动生成的 catch 块e.printStackTrace();}return str;}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}public Double nextDouble() {return Double.parseDouble(next());}public BigInteger nextBigInteger() {return new BigInteger(next());}}
}
