I will not include the problem in the blog anymore.

Here is the solution for this program problem. To simulate the circle, we can attach two necklace together to do the calculation. It is obvious we will double the time, but this make the result more intuitive to sense.

/*
ID:yuebing1
LANG:C++
TASK:beads
 */
#include<fstream>
#include<string>
#include<vector>
using namespace std;
int main()
{
	ifstream fin("beads.in");
	ofstream fout("beads.out");
	/**/
	int iNum;
	fin>>iNum;
	string sNecklace,sLinkedNecklace;
	fin>>sNecklace;
	sLinkedNecklace=sNecklace+sNecklace;//"+" operation for string.
	int color(0); //color is used to track the color of last bead
	int current(0);//current is used to track
	//the number of beads on current segment
	int previous(0);// previous is used track
	//the number of beads on previous segment
	int result(0);
	int w(0);
	for(int i=0;(i<iNum*2)&&(previous+current<iNum);++i)
	{
		if (sLinkedNecklace[i]=='w')
			{
				w++;
				++current;
			}
		else if (sLinkedNecklace[i]==color)
			{
				w=0;
				++current;
			}
		else
		{
			color=sLinkedNecklace[i];
			if(previous+current>result) result=previous+current;
			previous=current-w;
			current=1+w;
			w=0;
		}

	}
if(result<(current+previous))
		{
			fout<<current+previous<<endl;
		}
	else
		{
		fout<<result<<endl;
		}
}

Is Friday the 13th really an unusual event?

That is, does the 13th of the month land on a Friday less often than on any other day of the week? To answer this question, write a program that will compute the frequency that the 13th of each month lands on Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday over a given period of N years. The time period to test will be from January 1, 1900 to December 31, 1900+N-1 for a given number of years, N. N is non-negative and will not exceed 400.

There are few facts you need to know before you can solve this problem:

* January 1, 1900 was on a Monday.
* Thirty days has September, April, June, and November, all the rest have 31 except for February which has 28 except in leap years when it has 29.
* Every year evenly divisible by 4 is a leap year (1992 = 4*498 so 1992 will be a leap year, but the year 1990 is not a leap year)
* The rule above does not hold for century years. Century years divisible by 400 are leap years, all other are not. Thus, the century years 1700, 1800, 1900 and 2100 are not leap years, but 2000 is a leap year.

Do not use any built-in date functions in your computer language.

Don’t just precompute the answers, either, please.
my experience
the return value should be “return 0″

 /*
ID:yuebing1
 PROG:friday
 LANG:C++
 */
#include<fstream>
#include<vector>
using namespace std;
int main()
{	int iN;
		ifstream fin("friday.in");
		ofstream fout("friday.out");

		long day(13);	//bias for Monday,1900:1:1.
		int day_name(0);
		vector<int> statistics(7,0);
		int month[12]={31,0,31,30,31,30,31,31,30,31,30,31};
	fin>>iN;
	int limit=1900+iN;
	for(int i=1900; i
	<limit; i++)
	{
		if ((i%4==0)&&(i%100!=0)&#124;&#124;(i%400==0))//leap year
		{
			month[1]=29;
		}
		else	//normal year
		{
			month[1]=28;
		}

			for(int j=0;j<12;j++)
			{
			day_name=(day)%7;
			statistics[day_name]+=1;
			day+=month[j];
			}
	}
fout<<statistics[6];
	for(int i=0;i<=5;i++)
			fout<<' '<<statistics[i];
			fout<<endl;
	return 0;
}

There is another way of figuring out the date's name actually , in the following program you can see how it works.
/*
ID:yuebing1
PROB:friday
LANG:C++
 */
#include<iostream>
#include<fstream>
using namespace std;

int to(int x,int m,int d)
{
		if(m<3){m+=12;x--;}
		int c=(x/100);
		int y=x-100*c;
		int w=((c/4)-(2*c)+y+(y/4)+(26*(m + 1)/10)+d-1)%7;
		return (w+7)%7;
}

int main()
{
int n;
ifstream fin("friday.in");
ofstream fout("friday.out");
fin>>n;
	int x[7]={0};
for(int i=1900;i<1900+n;i++)
{
	for(int j=1;j<=12;j++)
	{
	x[to(i,j,13)]++;
	}
}

fout<<x[6];
for(int i=0;i<=5;i++)
	{
	fout <<" "<<x[i];
	}
	fout <<endl;
	return 0;
}

By comparing the difference between other's code and mine, I find that mine is not quite standard but more concise.
/*
ID: lixianm1
PROG: friday
LANG: C++
*/
#include
#include
#include
#include
#include
#include 

inline bool IsLeapYear(unsigned int nYear) {return ((!(nYear&#38;3))&#38;&#38;(0!=nYear%100))&#124;&#124;(0==nYear%400);}

int main(int argc, char* argv[])
{
    //////////////////////////////////////////////////////////////////////////open the file
    std::string strInFile    = "friday.in";
    std::string strOutFile    = "friday.out";

    std::ifstream fin(strInFile.c_str());
    std::ofstream fout(strOutFile.c_str());

    if (!fin)
    {
        std::cout&#60;&#60;"failed to open file for read"&#60;&#60;std::endl;
        return 1;
    }
    if (!fout)
    {
        std::cout&#60;&#60;"failed to open file for write"&#60;&#62;N;

    //////////////////////////////////////////////////////////////////////////process

    static int month_day[]={0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    static int month_day2[]={0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    int aWeekCount[7]={0};
    unsigned int nTotalDays= 0;

    for (int i=0; i&#60;N; ++i)
    {
        for (int nMonth=1; nMonth&#60;13; ++nMonth)
        {
            ++aWeekCount[(nTotalDays+12)%7];

            if (!IsLeapYear(1900+i))
            {
                nTotalDays+=month_day[nMonth];
            }else
            {
                nTotalDays+=month_day2[nMonth];
            }
        }
    }

    //////////////////////////////////////////////////////////////////////////write the process result
    for (int i=0; i&#60;6; ++i)
    {
        fout&#60;&#60;aWeekCount[(i+5)%7]&#60;&#60;" ";
    }
    fout&#60;&#60;aWeekCount[4]&#60;&#60;std::endl;

    //////////////////////////////////////////////////////////////////////////end and exit
    fin.close();
    fout.close();

    //system("pause");
    return 0;
}

At the end is the program from analysis which is written in c:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

int
isleap(int y)
{
    return y%4==0 && (y%100 != 0 &#124;&#124; y%400 == 0);
}

int mtab[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

/* return length of month m in year y */
int
mlen(int y, int m)
{
    if(m == 1)    /* february */
        return mtab[m]+isleap(y);
    else
        return mtab[m];
}

void
main(void)
{
    FILE *fin, *fout;
    int i, m, dow, n, y;
    int ndow[7];

    fin = fopen("friday.in", "r");
    fout = fopen("friday.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d", &n);

    for(i=0; i<7; i++)
        ndow[i] = 0;

    dow = 0;    /* day of week: January 13, 1900 was a Saturday = 0 */
    for(y=1900; y<1900+n; y++) {
        for(m=0; m<12; m++) {
            ndow[dow]++;
            dow = (dow+mlen(y, m)) % 7;
        }
    }

    for(i=0; i<7; i++) {
        if(i)
            fprintf(fout, " ");
        fprintf(fout, "%d", ndow[i]);
    }
    fprintf(fout, "\n");

    exit(0);
}

]]>



http://jiangyuebing.wordpress.com/2009/02/13/usaco-greedy-gift-giver/
Fri, 13 Feb 2009 20:44:26 +0000
jiangyuebing
http://jiangyuebing.wordpress.com/2009/02/13/usaco-greedy-gift-giver/

 A group of NP (2 ≤ NP ≤ 10) uniquely named friends has decided to exchange gifts of money. Each of these friends might or might not give some money to any or all of the other friends. Likewise, each friend might or might not receive money from any or all of the other friends. Your goal in this problem is to deduce how much more money each person gives than they receive.
The rules for gift-giving are potentially different than you might expect. Each person sets aside a certain amount of money to give and divides this money evenly among all those to whom he or she is giving a gift. No fractional money is available, so dividing 3 among 2 friends would be 1 each for the friends with 1 left over — that 1 left over stays in the giver’s “account”.
In any group of friends, some people are more giving than others (or at least may have more acquaintances) and some people have more money than others.
Given a group of friends, no one of whom has a name longer than 14 characters, the money each person in the group spends on gifts, and a (sub)list of friends to whom each person gives gifts, determine how much more (or less) each person in the group gives than they receive.
/*
ID:yuebing1
PROG:gift1
LANG:C++
*/
#include <fstream>
#include <map>
#include <string>
#include <vector>
#include <queue>
using namespace std;
int main()
{
	ifstream fin("gift1.in");
	ofstream fout("gift1.out");
	int iTotal;
	string name;
	map <string, int> people;
	int money,divide;
	map <string,int>::iterator person;
	queue <string> record;
	fin>>iTotal;

	for (int i=0;i<iTotal;i++)
	{
			fin>>name;
			people.insert(make_pair(name,0));
			record.push(name);
	}
	for(int i=0;i<iTotal;i++)
	{
		fin>>name;
		person=people.find(name);
		fin>>money>>divide;
	if(divide>0)
		person->second-=(money/divide)*divide;

		for(int j=0;j<divide;j++)
		{
			fin>>name;
			person=people.find(name);
			person->second+=money/divide;

		}

	}

	for(int i=0;i<iTotal;i++)
		{
			name=record.front();
			record.pop();
			person=people.find(name);
			fout<<name<<' '<<person->second<<endl;
		}
}

]]>



http://purplelight.wordpress.com/2009/02/11/usaco-february-contest-gold-division/
Wed, 11 Feb 2009 05:43:15 +0000
whatevalah
http://purplelight.wordpress.com/2009/02/11/usaco-february-contest-gold-division/

yeehaa… akhirnya setelah kontes pertama divisi Gold gue bulan lalu hancur lebur dengan hanya 162 poin (algo ngaco, salah baca soal, alhasil ngais2 nilai sekitar 1-3 test case per soal -_-”), untuk bulan ini ada sedikit kemajuan    Full score di 1 soal (prob: revamp), dan 1 soal lagi (shuttle) kena WA karena rekurens gue emang salah!! Jadinya cuma dapet beberapa case… Trus 1 lagi soal ga gue kerjain, itu soal paling keren lupa judulnya, yang jelas main2 saham punya, agak mirip knapsack tapi bingung gmn… wkwkwkwkwk…..
- Yang “revamp” gue pake a* tapi tanpa heuristik, cuma di bound dengan state-state. Awalnya sempet takut kena TLE, tapi setelah itung2 keknya pas… jadi hajar dah… 

- Yang shuttle cuma dapat yang DP o(nc), tapi ngaco banget… hahaha….
Here is the result :
- – - – - – - – - – - – - – - ss_imposs1 – - – - – - – - – - – - – - -

- – - – - – - – - – - – - – - USACO FEB09 – - – - – - – - – - – - – - – 
             Eko Mirhard  / IDN / Grad: 2007 / Div: Gold

As an observer, YOUR RESULTS MIGHT NOT APPEAR ON THE FINAL WEB LISTINGS
Below are the results of grading your submissions against the test data for

the contest.
============================== SUMMARY =============================
                            — case number –

            1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17

revamp      *  *  *  *  *  *  *  *  *  *  *

shuttle     *  *  *  x  x  x  x  x  x  x  x  x  x  x  x  x  x
. = no entry   t = time exceeded    *   = correct answer

x = wrong      s = signal           e   = bad exit status
Overall soal-soalnya sperti biasa keren2 dan cukup fresh… 

Bagi yang mau baca soal, ke sini aja, kalo mau unduh test case atau baca pembahasan, bisa ke sini
C u then… ^^
]]>