JAVA题解：

import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s = br.readLine().trim();
        String[] parts = br.readLine().trim().split(" ");
        int a = Integer.parseInt(parts[0]);
        int b = Integer.parseInt(parts[1]);
        
        solve(s, a, b);
    }
    
    public static void solve(String s, int a, int b) {
        int n = s.length();
        
        // 计算前缀余数 pref[i] = s[:i] mod a
        int[] pref = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            pref[i] = (pref[i - 1] * 10 + (s.charAt(i - 1) - '0')) % a;
        }
        
        // 计算后缀余数 suf[i] = s[i:] mod b
        int[] suf = new int[n + 1];
        long p = 1; // 当前位权
        for (int i = n - 1; i >= 0; i--) {
            suf[i] = (int)((s.charAt(i) - '0') * p + suf[i + 1]) % b;
            p = (p * 10) % b;
        }
        
        // 从右向左枚举切割点，找到第一个满足条件的切割点
        for (int i = n - 1; i > 0; i--) {
            // 后半部分不能有前导零
            if (s.charAt(i) == '0') {
                continue;
            }
            
            // 检查两部分是否分别能被a和b整除
            if (pref[i] == 0 && suf[i] == 0) {
                System.out.println("YES");
                System.out.println(s.substring(0, i));
                System.out.println(s.substring(i));
                return;
            }
        }
        
        // 没有找到合适的切割点
        System.out.println("NO");
    }
    
    // 用于处理大整数的辅助方法
    public static String[] solveWithBigInteger(String s, int a, int b) {
        int n = s.length();
        
        // 从右向左枚举切割点
        for (int i = n - 1; i > 0; i--) {
            // 后半部分不能有前导零
            if (s.charAt(i) == '0') {
                continue;
            }
            
            String left = s.substring(0, i);
            String right = s.substring(i);
            
            BigInteger leftNum = new BigInteger(left);
            BigInteger rightNum = new BigInteger(right);
            
            if (leftNum.mod(BigInteger.valueOf(a)).equals(BigInteger.ZERO) &&
                rightNum.mod(BigInteger.valueOf(b)).equals(BigInteger.ZERO)) {
                return new String[]{"YES", left, right};
            }
        }
        
        return new String[]{"NO"};
    }
}