/// Input - A Date, in the form DD/MM/YYYY. /// Output - Day of the Week on the date. #include #include #include #include #include #include #include using namespace std; using s = string; bool LeapYear(short y) { return (y%4==0&&y%100!=0)||(y%400==0); } short Simplify(short ods) { ods%=7; return ods>=0?ods:ods+7; } int f(double in) { return floor(in); } void DateToDay(s dt) { short yy,dd,mm,ody,odd,ods,od; short mth[12] = {31,28,31,30,31,30, 31,31,30,31,30,31}; short mcode[12] = {11,12,1,2,3,4, 5,6,7,8,9,10}; cout.fill('0'); replace(dt.begin(),dt.end(),'/',' '); stringstream ss(dt); ss>>dd>>mm>>yy; cout<12||dd>mth[mm-1]||dd<1||mm<1||yy==0) {cout<<" is not a valid date."; return; } if(mm<3) yy--; ody=yy/100; odd=yy%100; ods = dd + f((13*mcode[mm-1]-1)/5) + odd + f(odd/4) + f(ody/4) - 2*ody; od = Simplify(ods); if(yy+1<0) { cout<<" [BC]"; od--; if(od<0) od = 6; } if (od==0) cout<<" is a Sunday. \n"; else if(od==1) cout<<" is a Monday. \n"; else if(od==2) cout<<" is a Tuesday. \n"; else if(od==3) cout<<" is a Wednesday.\n"; else if(od==4) cout<<" is a Thursday. \n"; else if(od==5) cout<<" is a Friday. \n"; else if(od==6) cout<<" is a Saturday. \n"; } int main(void) { s d; regex pat(s("\\d{1,2}[/]\\d{1,2}")+ s("[/][-]?\\d{1,4}")); cin>>d; bool valid = regex_match(d,pat); if(d.empty()||!valid) d="01/01/0001"; DateToDay(d); }

Challenge - Date to Day of Week