Jasper

Tag: poj

poj_1009(TLE&MLE)

/*
//ACM Poj 1009
//File: 1009.cpp
//Author: Kangzj
//Mail: Kangzj@mail.bnu.edu.cn
//Date: April 29, 2009
*/
#include
#include
#include
#include $using namespace std;
struct node
{
  int pix;
  int count;
};
int main()
{
  int width,i;
  int pix;
  long len;
  ostringstream buff;
    while(cin>>width&&width!=0)
  {
    //读入图像并解码
    vector imgin;
    while(cin>>pix>>len&&len!=0)
    {
      vector temp(len,pix);
      imgin.insert(imgin.end(),temp.begin(),temp.end());
    }
    int height;
    int size=imgin.size();
    height=size/width;
    vector imgout(size,0);
    for(i=0;i =start)
        {
          max=abs(imgin[pos-1]-imgin[pos]);
        }
        if((pos+1) m?max:m;
        }
        if(pos+width m?max:m;
          if(pos+width+1 m?max:m;
          }
          if(pos+width-1>=start+width)
          {
            int m=abs(imgin[pos+width-1]-imgin[pos]);
            max=max>m?max:m;
          }
        }
        if(pos-width>=0)
        {
          int m=abs(imgin[pos-width]-imgin[pos]);
          max=max>m?max:m;
          if(pos-width+1< =start-1)
          {
            int m=abs(imgin[pos-width]-imgin[pos]);
            max=max>m?max:m;
          }
          if(pos-width-1>=start-width)
          {
            int m=abs(imgin[pos-width-1]-imgin[pos]);
            max=max>m?max:m;
          }
        }
        imgout[pos]=max;
      }
    }
    //将输出图像编码
    buff< rle;
    //第一个先放进去
    node tn;
    tn.pix=imgout[0];
    tn.count=0;
    rle.push_back(tn);
    //游程编码
    for(i=0;i 虽然AC了，但其实是不合格的，时间和空间都超了，500000000的数据量没法处理，好在测试集里没有这样的数据。。。转了一篇别人做的，质量真的不错，想到了好多我根本就没去想的问题，受教了。
代码完全是按题意做的，一点也没有改进：$

2009-04-30

poj_1008

/*
//ACM Poj 1008
//File: 1008.cpp
//Author: Kangzj
//Mail: Kangzj@mail.bnu.edu.cn
//Date: April 29, 2009
*/
#include
#include
using namespace std;
struct TYD
{
	int Dnum;
	string Dname;
};
int main()
{
    string TkDays[20]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};
	string HbMons[19]={"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu"};
	ostringstream buff;
	TYD Tday;
	int i,j,n;
	int Hyear,Hday, Tyear;
	string Hdaystr, Hmonth;
	int days,Tdays;
	cin>>n;
	buff< >Hdaystr>>Hmonth>>Hyear;
		Hdaystr=Hdaystr.substr(0,Hdaystr.size()-1);
		if (Hdaystr.size()==2)
		{
			Hday=(Hdaystr[0]-'0')*10+(Hdaystr[1]-'0');
		}else
		{
			Hday=(Hdaystr[0]-'0');
		}
		if(Hmonth=="uayet")
		{
			j=18;
		}
		else
		{
			for(j=0;j<18;j++)
			{
				if(Hmonth==HbMons[j])
				{
					break;
				}
			}
		}
		days=Hyear*365+j*20+Hday;
		Tyear=days/260;
		Tdays=days%260;//注意边界！
		Tday.Dname=TkDays[Tdays%20];
		Tday.Dnum=Tdays%13+1;
		buff<

2009-04-29

poj_1007

/*
//ACM Poj 1007
//File: 1007.cpp
//Author: Kangzj
//Mail: Kangzj@mail.bnu.edu.cn
//Date: April 28, 2009
*/
#include
#include
using namespace std;
struct revNum
{
  int rvNums;
  int order;
};
int getReverseNum(string DNA, int n)
{
  int *count=new int[n];
  int all=0,i;
  for (i=0;i=0;i--)
  {
    for(int j=0;j>n>>m;
  string *DNA=new string[m];
  int count=0,temp,i;
  revNum *rvNum=new revNum[m];
  for(i=0;i>DNA[i];
    temp=getReverseNum(DNA[i],n);
    //插入排序
    if(count==0)
    {
      rvNum[0].rvNums=temp;
      rvNum[0].order=i;
      count++;
    }
    else
    {
            int j,flag=0;
      for(j=0;jj;k--)
          {
            rvNum[k]=rvNum[k-1];
          }
          rvNum[j].rvNums=temp;
          rvNum[j].order=i;
          count++;
          flag=1;
          break;
        }
      }//for(int j=0;j

2009-04-28

poj_1005

/*
//ACM Poj 1005
//File: 1005.cpp
//Author: Kangzj
//Mail: Kangzj@mail.bnu.edu.cn
//Date: April 28, 2009
*/
#include
#include $#define PI 3.14159265358979
using namespace std;
int main()
{
  double distance;
  int n;
  cin>>n;
  double *x=new double[n];
  double *y=new double[n];
  int *year=new int[n];
  for(int i=0;i >x[i]>>y[i];
    distance=sqrt(x[i]*x[i]+y[i]*y[i]);
    int j=1;
    double r=0;
    while(1)
    {
      r=sqrt(100/PI+r*r);
      if(distance< =r)break;
      j++;
    }
    year[i]=j;
  }
  for(int i=0;i$

2009-04-28

poj_2000

/*
//ACM Poj 2000
//File: 2000.cpp
//Author: Kangzj
//Mail: Kangzj@mail.bnu.edu.cn
//Date: April 24, 2009
*/
#include
#include
#include $using namespace std;
int main()
{
  int day;
  int coinDay, coinAll;
  ostringstream buff;
    while(1)
  {
        cin>>day;
        if(day==0)break;
    float temp=sqrt(1+8*day);
    int tmp=(int)temp;
    if(temp!=(float)tmp && tmp%2!=0)
    {
      tmp+=2;
    }else if(temp!=(float)tmp && tmp%2==0)
    {
      tmp++;
    }else
    {
      tmp=tmp;
    }
    coinDay=(-1+tmp)/2;
    coinDay--;
    coinAll=coinDay*(coinDay+1)*(2*coinDay+1)/6;
    coinAll+=(coinDay+1)*(day-(coinDay+1)*coinDay/2);
    buff< 需要选择G++才能通过，如果想用C++，只要加几个强制转换就可以了。$

2009-04-28

poj_1006

Problem: http://poj.grids.cn/problem?id=1006
Solution:

#include
#include
#include
using namespace std;
int main()
{
    int p, e, i, n=0;
    long cp, ce, ci;
    short d;
    int j=1;//计数用
    ostringstream buf;
    while(1)
    {
        cin>>p>>e>>i>>d;
        if(p==-1 && e==-1 && i==-1 && d==-1)
        {
            break;
        }
        n=d;
        while(1)
        {
            n++;
            if((n-p)%23==0 && (n-e)%28==0 && (n-i)%33==0)
            {
                n-=d;
                if(n==0)
                {
                    n=d;
                    continue;
                }
                break;
            }
            //cout< <"*";
        }
        buf<<"Case "<

2009-04-16

poj_1004

#include
using namespace std;
int main()
{
    int i;
    float all=0, temp=0;
    for(i=0;i<12;i++)
    {
        cin>>temp;
        all+=temp;
    }
    all+=0.005;
    cout.precision(2);
    //cout.setf(ios_base::showpoint);
    cout< <"$"<

2009-04-16