BigInteger target = new BigInteger("2828718396");
        System.out.println(target.toString(16));
        System.out.println(Long.toHexString(2828718396l));

------解决方案--------------------

Java code

    public static void main(String[] args) {
        int a = 527461410;
        int a1 = (a & 0xff);
        int a2 = ((a >> 8) & 0x00ff);
        int a3 = ((a >> 16) & 0x00ff);
        int a4 = ((a >> 24) & 0x00ff);
        System.out.println(byteToHex(a1) + byteToHex(a2) +byteToHex(a3) +byteToHex(a4));
    }

    /**
     * byte转换为十六进制字符串
     * 
     * @param b
     *            byte
     * @return 转换后十六进制字符串
     */
    public static String byteToHex(int b) {
        char b01 = getChar((b & 0xf0) >> 4);
        char b02 = getChar(b & 0x0f);
        return String.valueOf(b01) + String.valueOf(b02);
    }
    
    /**
     * 取得16进制字符串
     * 
     * @param b
     *            16进制的数字表示
     * @return 16进制的char
     * @throws IllegalArgumentException
     *             如果非16进制的数字则抛出该异常
     */
    public static char getChar(int b) {
        if (b >= 0 && b <= 9) {
            return (char) ('0' + b);
        }

        if (b >= 10 && b <= 15) {
            return (char) ('A' - 10 + b);
        }

        throw new IllegalArgumentException("error byte:" + b);
    }

------解决方案--------------------
楼主，你的10进制数和16进制数不相等，那你的真实意图是什么呢？
正常Java中10进制和16进制转换是比较简单的，如
		long b = Long.parseLong("2828718396");
		System.out.println(Long.toHexString(b));
不过结果应该是
a89ad13c