那些年，我寫程式的日子

c++ Uva 11450

問題連結：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2445

DP問題=>背包變型

#include<iostream>
using namespace std;
int main()
{
int n,m,c,k,min,sum,max;
int table[20][21];
int T[21][201];
cin>>n;
for(int i=1;i<=n;i++)
{
sum=0;
cin>>m;
cin>>c;
for(int j=0;j<=c-1;j++)
{
min=99999;
cin>>table[j][0];
k=1;
while(k<=table[j][0])
{
cin>>table[j][k];
if(table[j][k]<=min)
min=table[j][k];
k++;
}
sum=sum+min;
}
if(sum>m)
{
cout<<"no solution"<<endl;
continue;
}
else
{
for(int b=0;b<=c;b++)
for(int a=0;a<=m;a++)
T[b][a]=0;
for(int q=1;q<=c;q++)
for(int p=1;p<=m;p++)
{
for(int o=1;o<=table[q-1][0];o++)
{
if(p>=table[q-1][o])
{
if((q==1) && (T[q-1][p-table[q-1][o]]+table[q-1][o])>=T[q][p])
T[q][p]=T[q-1][p-table[q-1][o]]+table[q-1][o];
else
if((T[q-1][p-table[q-1][o]]+table[q-1][o])>=T[q][p] && (T[q-1][p-table[q-1][o]])>0)
T[q][p]=T[q-1][p-table[q-1][o]]+table[q-1][o];
}
}

}
cout<<T[c][m]<<endl;
}
}
return 0;
}

c++ 大數加法

#include<iostream>

#include<string>

#include<vector>

using namespace std;

int main()

{

vector<int>bit1,bit2;

string input1,input2;

int length1,length2,tmp;

bool flag=0,next=0; //預先判斷要不要進位；下一次進算時把進位的數字加上去

cout<<"輸入第一個數字"<<endl;

cin>>input1;

cout<<"輸入第二個數字"<<endl;

cin>>input2;

for(int i=0;i<=input1.length()-1;i++) //把string的char一個一個存成int

bit1.push_back(input1[i]-'0');

for(int j=0;j<=input2.length()-1;j++)

bit2.push_back(input2[j]-'0');

length1=input1.length()-1;

length2=input2.length()-1;

if(length1>=length2) //當第1個字串大於第2個字串時

while(1)

{

if(length2==-1)

{

bit1[length1]=(bit1[length1]+next)%10;

if(next==1)

{

bit1.push_back(next);

for(int k=0;k<=input1.length();k++)

{

tmp=bit1[k];

bit1[k]=bit1[input1.length()];

bit1[input1.length()]=tmp;

}

break;

}

if((bit1[length1]+bit2[length2])>=10)

flag=1;

bit1[length1]=(bit1[length1]+bit2[length2])%10+next;

next=flag;

flag=0;

length1--;

length2--;

}

else //當第2個字串大於第1個字串時

while(1)

{

if(length1==-1)

{

bit2[length2]=(bit2[length2]+next)%10;

if(next==1)

{

bit2.push_back(next);

for(int k=0;k<=input2.length();k++)

{

tmp=bit2[k];

bit2[k]=bit2[input2.length()];

bit2[input2.length()]=tmp;

}

break;

}

if((bit1[length1]+bit2[length2])>=10)

flag=1;

bit2[length2]=(bit1[length1]+bit2[length2])%10+next;

next=flag;

flag=0;

length1--;

length2--;

}

if(input1.length() >= input2.length())

{

if(next==1)

for(int i=0;i<=input1.length();i++)

cout<<bit1[i];

else

for(int i=0;i<=input1.length()-1;i++)

cout<<bit1[i];

}

else

{

if(next==1)

for(int i=0;i<=input1.length();i++)

cout<<bit1[i];

else

for(int i=0;i<=input1.length()-1;i++)

cout<<bit1[i];

}

cout<<endl;

return 0;

}

c++ 簡易linking list

#include<iostream>

using namespace std;

struct link

{

int number;

link *next;

};

int main()

{

int times;

link first;

link *current;

cout<<"輸入要多大的int空間：";

cin>>times;

cout<<"請輸入數值"<<endl;

cin>>first.number;

first.next=new link;

current=first.next;

for(int i=1;i<times;i++)

{

cin>>current->number;

current->next=new link;

current=current->next;

}

current=&first;

for(int j=0;j<times;j++)

{

cout<<"第"<<j+1<<"次輸入的數值"<<current->number<<endl;

current=current->next;

}

return 0;

}

c++ 一串string type的數字，以逗號分開存成int的type

#include<iostream>
#include<vector>
#include<string>
using namespace std;
int main()
{
vector<int>i_num;
string str="10,20,30,40,50";
string num="";
int location=-1;
for(int i=0;i<=str.length()-1;i++)
{
if(str[i]==',')
{
for(int j=location+1;j<=i-1;j++)
num=num+str[j];
i_num.push_back(atoi(num.c_str()));
num="";
location=i;
}
if(i==str.length()-1)
{
for(int k=location+1;k<=i;k++)
num=num+str[k];
i_num.push_back(atoi(num.c_str()));
num="";
location=i;
}
}
for(int p=0;p<=4;p++)
cout<<i_num[p]<<endl;
return 0;
}

c++ triangle

這是執行檔案，給大家下載去玩
https://skydrive.live.com/redir.aspx?cid=a4872f58128956b2&resid=A4872F58128956B2!268&parid=A4872F58128956B2!267&authkey=!APmGHg_J5c00EGw

下面則是列出MFC主要增加的部分

class CtriangleDlg : public CDialogEx //全域變數抓輸入資料
{
public:
int number;
bool start;
//其他略
};

BOOL CtriangleDlg::OnInitDialog() //初始化痊癒變數的數值
{
start=0;
//其他略
}

void CtriangleDlg::OnPaint()
{
if (IsIconic())
{
//略
}
else
{
CPaintDC pDC(this);
if(start==1) //button按下時開始
{
pDC.MoveTo(100,500); //初始的三點座標
pDC.LineTo(300,100);
pDC.LineTo(500,500);
pDC.LineTo(100,500);
if(number==0)
;
else
triangle(100,500,300,100,500,500,0,number,&pDC);
}
CDialogEx::OnPaint();
}
}

void triangle(double x1,double y1,double x2,double y2,double x3,double y3,int count,int N,CPaintDC* draw)
{
double leftx=(x1+x2)/2;
double lefty=(y1+y2)/2;
double rightx=(x1+x3)/2;
double righty=(y1+y3)/2;
double downx=(x2+x3)/2;
double downy=(y2+y3)/2;
draw->MoveTo(int(leftx),int(lefty));
draw->LineTo(int(rightx),int(righty));
draw->LineTo(int(downx),int(downy));
draw->LineTo(int(leftx),int(lefty));
count++;
if(count<N)
{
triangle(leftx,lefty,x2,y2,downx,downy,count,N,draw);
triangle(x1,y1,leftx,lefty,rightx,righty,count,N,draw);
triangle(rightx,righty,downx,downy,x3,y3,count,N,draw);
}
}

void CtriangleDlg::OnBnClickedButton1()
{
CString tmp;
GetDlgItemText(IDC_EDIT1,tmp); //抓取edit box的資料
number= _ttoi(tmp.GetBuffer(0));
start=1;
Invalidate(); //畫面清空重新畫圖
}

NCNU LISP hand write 1

(define (A x y )

(cond((= y 0) 0)

((= x 0) (* 2 y))

((= y 1) 2)

(else (A (- x 1) (A x (- y 1 )))))

)

(define (h n) (A 2 n ) )的結果&證明

NCNU LISP HW3

exercise 3: linear recursive and linear iterative
but the function is changed:

            f(n) = n                        if n < 3 
                   2*f(n-1)+f(n-2)-f(n-3)   otherwise

(define (recursive x)
(cond ((= x 0) 0)
((= x 1) 1)
((= x 2) 2)
(else (- (+ (* (recursive (- x 1)) 2) (recursive (- x 2)))
(recursive (- x 3))))
)
)
(define (interative x)
(cond ((= x 0) 0)
((= x 1) 1)
((= x 2) 2)
(else (do-interative 0 1 2 (- x 2)))
)
)
(define (do-interative a b c x)
(cond ((= x 0) c)
(else (do-interative b c (- (+ (* c 2) b) a) (- x 1)))
)
)

2013年3月29日 星期五

2013年3月29日星期五