黑色星期五（Friday the thirteenth）的程序。這些程序其實都很簡單，沒有很複雜的算法在裏面，追求極致的Geek們可以從代碼的長度、優雅簡潔等方面下功夫了，俺僞Geek一個，寫了個程序能通過，就不糾結了，哈哈…
第二題有點小麻手，關鍵是將兩條鏈子拼湊在一起，Problem Description的最後也提示了這一點。
/*
ID: fairyroad
PROG: friday
LANG: C++
*/
#include <iostream>
#include <fstream>
using namespace std;
inline bool IsLeapyear(int year)
{
return (year%4==0 && year%100!=0 || year%400==0) ? true : false;
}
int Leapyear[12]={31,29,31,30,31,30,31,31,30,31,30,31};
int Averageyear[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int result[7];
int date=0; // the day associated with 13-th of the month
int year=1900;
int preSurplus=0; // number of days left from 13-th of last month
int main()
{
ofstream fout ("friday.out");
ifstream fin ("friday.in");
int n;
fin>>n;
for(int i=0;i<n;i++)
{
bool flag=IsLeapyear(year);
for(int j=0;j<12;j++)
{
date=((preSurplus+13)%7+date)%7;
++result[date];
preSurplus=flag?Leapyear[j]-13:Averageyear[j]-13;
}
++year;
}
fout<<result[6]<<" "; //Saturday
fout<<result[0]<<" "; //Sunday
fout<<result[1]<<" ";
fout<<result[2]<<" ";
fout<<result[3]<<" ";
fout<<result[4]<<" ";
fout<<result[5]<<endl;
return 0;
}
/*
ID: fairyroad
LANG:C++
TASK:beads
*/
#include<fstream>
#include<string>
#include<vector>
using namespace std;
int main()
{
ifstream fin("beads.in");
ofstream fout("beads.out");
int len;
fin>>len;
string str,myStr;
fin>>str;
myStr=str+str;
int color=0;
int patrol=0;
int header=0;
int max(0);
int wconut(0);
for(int i=0;(i<len*2)&&(header+patrol<len);++i)
{
if (myStr[i]=='w'){
wconut++;
++patrol;
}
else if (myStr[i]==color){
wconut=0;
++patrol;
}
else
{
color=myStr[i];
if(header+patrol>max) max=header+patrol;
header=patrol-wconut;
patrol=1+wconut;
wconut=0;
}
}
if(max<(patrol+header)) fout<<patrol+header<<endl;
else fout<<max<<endl;
}