public class Test {
    public static int f(int n) {
        if(n<3) {
            return 0;
        } 
        int count = 0;
        String s = String.valueOf(n);
        int index = 0;
        while(true){
            index = s.indexOf("3", index);
            if(index==-1){
                break;
            }
            index++;
            count++;
        }
        return f(n-1) + count;
    }
    
    public static void main(String args[]) {
        System.out.println(f(500));
    }
}

------解决方案--------------------
第一题貌似看到过：http://eastsun.javaeye.com/blog/59836
第二题也看到过用ArrayList拼出来一个
Java code

public class Test{
    public static native int invoke(int arg);

    static{
        System.load("/home/Estelle/Workspace/Other/libTest.so");
    }

    public static voide res

------解决方案--------------------
第一题
算法效率不高
大家一起研究
import java.io.*;


public class Count3 {
	private int count=0;
	
	public static void main(String[] args) {
		int n=0;
		BufferedReader input=new BufferedReader(new InputStreamReader(System.in));
		System.out.println("请输入一个整数N：输入0退出");
		try {
			n=Integer.parseInt(input.readLine());
		} catch (NumberFormatException e) {			
			e.printStackTrace();
		} catch (IOException e) {			
			e.printStackTrace();
		}
		Count3 count3=new Count3();
		
		while(n>0){
			System.out.println("1到"+n+"之间出现3的次数为："+count3.count(n));
			System.out.println("输入一个整数N：输入0退出");
			try {
				n=Integer.parseInt(input.readLine());
			} catch (NumberFormatException e) {
				e.printStackTrace();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		System.out.println("已退出");
	}
	
	public  int count(int n){
		int temp,k;
		count=0;
		for(int i=2;i<=n;i++){
			if(i%10==3){
				count++;
			}
			k=i;
			while((temp=k/10)!=0){
				if(temp%10==3){
					count++;
				}
				k=temp;
			}
		}
		return count;
		
	}
	

}