下面l出我的解法Q几乎可以一步到位求出来
实现代码如下Q?br />/**
*Author: Koth (http://weibo.com/yovn)
*Date: 2011-12-01
*/
#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>
int solve(int Y,int& X){
int m=0;
int t=Y;
if(Y<=0){
X=Y;
return 1;
}
while((t&1)==0){
m+=1;
t=t>>1;
}
if(m==32){
X=Y;
return 1;
}
int lastK=32;
for(;lastK>m+1;lastK--){
if(Y &(1<<(lastK-1))){
break;
}
}
//its a number st. exp(2,K)
if(lastK==(m+1)){
X=Y;
return 1;
}
int k=1<<(m+1);
int b=(Y>>m)-(1<<(lastK-m-1));
X=(1<<(lastK-m-2))+(b+1-k)/2;
if(X<=0){
k=k-1-((0-X)<<1);
X=0-X+1;
}
return k;
}
int main(int argc,char* argv[]){
if(argc<=1){
fprintf(stdout,"Usage:%s number\n",argv[0]);
return 0;
}
int Y=atoi(argv[1]);
int X=0;
int k=solve(Y,X);
fprintf(stdout,"%d=",Y);
for(int i=0;i<k;i++){
fprintf(stdout,"%d",X+i);
if(i<(k-1)){
fprintf(stdout,"+");
}
}
fprintf(stdout,"\n");
return 0;
}
*Author: Koth (http://weibo.com/yovn)
*Date: 2011-12-01
*/
#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>
int solve(int Y,int& X){
int m=0;
int t=Y;
if(Y<=0){
X=Y;
return 1;
}
while((t&1)==0){
m+=1;
t=t>>1;
}
if(m==32){
X=Y;
return 1;
}
int lastK=32;
for(;lastK>m+1;lastK--){
if(Y &(1<<(lastK-1))){
break;
}
}
//its a number st. exp(2,K)
if(lastK==(m+1)){
X=Y;
return 1;
}
int k=1<<(m+1);
int b=(Y>>m)-(1<<(lastK-m-1));
X=(1<<(lastK-m-2))+(b+1-k)/2;
if(X<=0){
k=k-1-((0-X)<<1);
X=0-X+1;
}
return k;
}
int main(int argc,char* argv[]){
if(argc<=1){
fprintf(stdout,"Usage:%s number\n",argv[0]);
return 0;
}
int Y=atoi(argv[1]);
int X=0;
int k=solve(Y,X);
fprintf(stdout,"%d=",Y);
for(int i=0;i<k;i++){
fprintf(stdout,"%d",X+i);
if(i<(k-1)){
fprintf(stdout,"+");
}
}
fprintf(stdout,"\n");
return 0;
}
]]>
证明Q?br /> 首先我们证明若N是素敎ͼ那么{式成立Q对于N=2,q是很明昄。以下证明N>2 的情形?br /> 1Q若N是素敎ͼ那么关于N同余的乘法群G={1,2,3....N-1}
G每个元素都有逆元Q映?f:a -> a^-1 Qf(a)=a^-1 是一个一一映射。现在，d一个元素，它的逆元要么是其它一个元素，或者是它本w?我们假设其中元素x的逆元是它本nQ那么x*x ≡1(mod N) =>(x+1)*(x-1)=K*N,而N是素敎ͼ所以要么x=N-1,要么x=1。也是_除了q两个元素，其它的元素的逆元都是映射到别的元素的。而N>2,是奇敎ͼ所以元素共有N-1个，也就是偶C元素。这P除了1和N-1外剩余的N-3个元素刚好结成两两一?x,y)(逆元是唯一的）使得f(x)=y=x^-1,也就是xy≡1(mod N).
现在把G的元素全部乘hQ让互ؓ逆元的元素组成一寚w?N-1)!=1*(N-1)*(x1*y1)*(x2*y2)...(xk*yk)≡1*(N-1)*1*1...1 (mod N)≡-1(mod N).
q样Q我们证明了一个方向了Q下面我们证明另一个方?br />
2Q若(N-1)! ≡ -1(mod N)成立Q则N是素敎ͼ若不Ӟ?N是和数。则(N-1)!肯定整除N,因ؓN的每个因子p满p>1,p<N?br /> 现在?N-1)!=K1*N.?N-1)! ≡ -1(mod N) => K1*N ≡ -1(mod N),由同余性质知，存在K2使得K1*N+1=K2*N,两边同时除以N得K1+1/N=K2.昄Q这是不可能的，所以若(N-1)! ≡ -1(mod N)成立Q则N是素?br />
证明完毕Q?br />
q里用群的概念能够简化一定的描述Q其实可以完全不用群的概늚Q只不过q样一来，描述更长点，J琐点！
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*
*/
import java.util.Stack;
import java.util.Vector;
/**
* @author yovn
*
*/
public class TreeDemo {
static interface NodeVistor
{
<T> void visit(BinaryTreeNode<T> node);
}
static class BinaryTree<T>
{
BinaryTreeNode<T> root;
public BinaryTree(BinaryTreeNode<T> root) {
this.root=root;
}
//no recursion ,pre-order
void NLRVisit(NodeVistor visitor)
{
BinaryTreeNode<T> pointer=root;
Stack<BinaryTreeNode<T>> stack=new Stack<BinaryTreeNode<T>>();
while(!stack.isEmpty()||pointer!=null)
{
if(pointer!=null)
{
visitor.visit(pointer);
if(pointer.rightChild!=null)
{
stack.push(pointer.rightChild);
}
pointer=pointer.leftChild;
}
//go to right child
else
{
pointer=stack.pop();
}
}
}
//no recursion , in-order
void LNRVisit(NodeVistor visitor)
{
BinaryTreeNode<T> pointer=root;
Stack<BinaryTreeNode<T>> stack=new Stack<BinaryTreeNode<T>>();
while(!stack.isEmpty()||pointer!=null)
{
if(pointer!=null)
{
stack.push(pointer);
pointer=pointer.leftChild;
}
//no left child here, then visit root and then go to right child
else
{
pointer=stack.pop();
visitor.visit(pointer);
pointer=pointer.rightChild;
}
}
}
//no recursion ,post-order,this one is the most complex one.
//we need know from which side ,it back(left or right)
void LRNVisit(NodeVistor visitor)
{
if(root==null)return;
BinaryTreeNode<T> pointer=root;
Stack<BinaryTreeNode<T>> stack=new Stack<BinaryTreeNode<T>>();
while(true)
{
//mark left
while(pointer!=null)
{
stack.push(pointer);
pointer=pointer.leftChild;
}
pointer=stack.pop();
while(pointer.visitedRight)//back from right child, so we can visit it now.
{
visitor.visit(pointer);
if(stack.isEmpty())return;
pointer=stack.pop();
}
pointer.visitedRight=true;
stack.push(pointer);
pointer=pointer.rightChild;
}
}
void levelOrder(NodeVistor visitor)
{
if(root==null)return;
BinaryTreeNode<T> pointer=root;
Vector<BinaryTreeNode<T>> queue=new Vector<BinaryTreeNode<T>>();
queue.add(pointer);
while(!queue.isEmpty())
{
BinaryTreeNode<T> node=queue.remove(0);
visitor.visit(node);
if(node.leftChild!=null)
{
queue.add(node.leftChild);
}
if(node.rightChild!=null)
{
queue.add(node.rightChild);
}
}
}
}
static class BinaryTreeNode<T>
{
BinaryTreeNode(T data)
{
this.data=data;
}
T data;
boolean visitedRight;
BinaryTreeNode<T> leftChild;
BinaryTreeNode<T> rightChild;
}
/**
*
*/
public TreeDemo() {
// TODO Auto-generated constructor stub
}
/**
* @param args
*/
public static void main(String[] args) {
BinaryTreeNode<String> root=new BinaryTreeNode<String>("A");
root.leftChild=new BinaryTreeNode<String>("B");
root.rightChild=new BinaryTreeNode<String>("C");
root.leftChild.leftChild=new BinaryTreeNode<String>("D");
root.rightChild.leftChild=new BinaryTreeNode<String>("E");
root.rightChild.rightChild=new BinaryTreeNode<String>("F");
root.rightChild.leftChild.rightChild=new BinaryTreeNode<String>("G");
root.rightChild.rightChild.leftChild=new BinaryTreeNode<String>("H");
root.rightChild.rightChild.rightChild=new BinaryTreeNode<String>("I");
NodeVistor visitor=new NodeVistor()
{
@Override
public <T> void visit(BinaryTreeNode<T> node) {
System.out.print("'"+node.data+"'");
}
};
BinaryTree<String> tree=new BinaryTree<String>(root);
System.out.println("pre-order visit");
tree.NLRVisit(visitor);
System.out.println();
System.out.println("in-order visit");
tree.LNRVisit(visitor);
System.out.println();
System.out.println("post-order visit");
tree.LRNVisit(visitor);
System.out.println();
System.out.println("level-order visit");
tree.levelOrder(visitor);
}
}
]]>
q个版本调整了一下Loader的机Ӟ保持EXE跟类文g然一体的同时又可以gq加载class.
另外q个版本的加密强度提高很多，用来代替那些classh软g应该也不错。：Q?nbsp; 阅读全文
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* Demonstrate KMP algorithm in Java
*
*
*/
public class KMP {
public static int indexOf(String target,String pattern)
{
int pLen=pattern.length();
int tLen=target.length();
//the fail function
int failFunc[]=new int[pLen];
failFunc[0]=-1;
//build fail function
for(int i=1;i<pLen;i++)
{
int j=failFunc[i-1];
while(pattern.charAt(i)!=pattern.charAt(j+1)&&j>=0)
{
//recursion
j=failFunc[j];
}
if(pattern.charAt(i)==pattern.charAt(j+1))
{
failFunc[i]=j+1;
}
else
{
failFunc[i]=-1;
}
}
int pPos=0,tPos=0;
while(tPos<tLen&&pPos<pLen)
{
if(target.charAt(tPos)==pattern.charAt(pPos))
{
//match ,then do forward
tPos++;
pPos++;
}
else if(pPos==0)
{
//target go forward
tPos++;
}
else
{
//target postion don't change,pattern go back
pPos=failFunc[pPos-1]+1;
}
}
if(pPos<pLen)return -1;
else return tPos-pLen;
}
}
]]>
3)gcc args:
gcc -Wall -O -fomit-frame-pointer -nostdinc -fno-builtin
4)link.ld
ld -T link.ld
OUTPUT_FORMAT("binary")
ENTRY(start)
phys = 0x00100000;
SECTIONS
{
.text phys : AT(phys) {
code = .;
*(.text)
*(.rodata)
. = ALIGN(4096);
}
.data : AT(phys + (data - code))
{
data = .;
*(.data)
. = ALIGN(4096);
}
.bss : AT(phys + (bss - code))
{
bss = .;
*(.bss)
. = ALIGN(4096);
}
end = .;
}
5)Bootable CDENTRY(start)
phys = 0x00100000;
SECTIONS
{
.text phys : AT(phys) {
code = .;
*(.text)
*(.rodata)
. = ALIGN(4096);
}
.data : AT(phys + (data - code))
{
data = .;
*(.data)
. = ALIGN(4096);
}
.bss : AT(phys + (bss - code))
{
bss = .;
*(.bss)
. = ALIGN(4096);
}
end = .;
}
mkisofs -R -b boot/grub/eltorito.s2 -no-emul-boot -boot-load-size 4 -boot-info-table -o yovnos.iso iso
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