• Source 
    1. #include<bits/stdc++.h>
    2. using namespace std;
    3.  
    4. string days(int y)
    5.  
    6. {
    7.         int c=0,d;
    8.         int i;
    9.  
    10.        if(y>2001)
    11.     {
    12.             for(i=2001;i<y;i++)
    13.             {
    14.             if((i%4==0 &&i%100!=0)||i%400==0)
    15.             c=c+2;
    16.             else
    17.             c++;
    18.             }
    19.             c=(1+(c%7))%7;
    20.     }
    21.     else
    22.     {
    23.             for(i=y;i<2001;i++)
    24.             {
    25.             if((i%4==0 &&i%100!=0)||i%400==0)
    26.             c+=2;
    27.             else
    28.             c++;
    29.             }
    30.             c=(8-(c%7))%7;
    31.     }
    32.                 string st="";
    33.                 if(c==2)
    34.                 st += "tuesday";
    35.                 if(c==3)
    36.                 st += "wednesday";
    37.                 if(c==4)
    38.                 st += "thursday";
    39.                 if(c==5)
    40.                 st += "friday";
    41.                 if(c==6)
    42.                 st += "saturday";
    43.                 if(c==0)
    44.                 st += "sunday";
    45.                 if(c==1)
    46.                 st += "monday";
    47. return st; 
    48.  
    49. }
    50.  
    51. int main()
    52. {
    53. // MAIN FORMULA - (year code + month code + century code + Date number - Leap year code)%7
    54. // Here as Date is 1 & Month is January hence DATE NUMBER is 1 and MONTH CODE is 0
    55. // If leap year SUBSTRACT 1 ELSE DON'T
    56. // Year code formula - (YY +(YY/4))%7 where in 1898 YY is 98
    57. // Century code - 1800s - 2; 1900s - 0; 2000s - 6; 2100s - 4; 2200s - 2; 2300s - 0;
    58. int number_of_years, year_code, month_code, yy, century_code, date_number, leap_year, year[1000], day, i, j;
    59. month_code = 0;
    60. date_number = 1;
    61. //scanf("%d",&number_of_years);
    62. for(i = 1800; i<=2600; i++)
    63. {
    64.     //printf(" * ");
    65.     year[i-1900] = i;
    66. }
    67.  
    68. for(j = 0; j<=700; j++)
    69. {  
    70.     year[j] = j+1800;
    71.     yy = 0; century_code = 0; year_code = 0; leap_year = 0; day = 0;
    72.    {
    73.         if(year[j]>=1800 && year[j]<=1899)
    74.         {century_code = 2; yy = year[j] - 1800; year_code = ((yy+(yy/4))%7);}
    75.         else if(year[j]>=1900 && year[j]<=1999)
    76.         {century_code = 0; yy = year[j] - 1900; year_code = ((yy+(yy/4))%7);}
    77.         else if(year[j]>=2000 && year[j]<=2099)
    78.         {century_code = 6; yy = year[j] - 2000; year_code = ((yy+(yy/4))%7);}
    79.         else if(year[j]>=2100 && year[j]<=2199)
    80.         {century_code = 4; yy = year[j] - 2100; year_code = ((yy+(yy/4))%7);}
    81.         else if(year[j]>=2200 && year[j]<=2299)
    82.         {century_code = 2; yy = year[j] - 2200; year_code = ((yy+(yy/4))%7);}
    83.         else if(year[j]>=2300 && year[j]<=2399)
    84.         {century_code = 0; yy = year[j] - 2300; year_code = ((yy+(yy/4))%7);}
    85.         else if(year[j]>=2400 && year[j]<=2499)
    86.         {century_code = 6; yy = year[j] - 2400; year_code = ((yy+(yy/4))%7);}
    87.         else if(year[j]>=2500 && year[j]<=2599)
    88.         {century_code = 4; yy = year[j] - 2500; year_code = ((yy+(yy/4))%7);}
    89.    }
    90.         if((year[j]%4 == 0) && (year[j]%100 != 0))
    91.         { leap_year = 1;}
    92.         else if(year[j]%400 == 0)
    93.         { leap_year = 1;}
    94.  
    95.    day = ((year_code +  month_code + century_code + date_number - leap_year)%7);
    96.     string st="";
    97.     switch(day)
    98.      {  case 0: st = "sunday";
    99.                     break ;
    100.         case 1: st += "monday";
    101.                   break ;
    102.         case 2: st += "tuesday";
    103.                     break ;
    104.         case 3: st += "wednesday";
    105.                     break ;
    106.         case 4: st += "thursday";
    107.                     break ;
    108.         case 5: st += "friday" ;
    109.                     break ;
    110.         case 6: st += "saturday";
    111.                     break ;
    112.       }
    113.       string temp = days(year[j]);
    114.       //printf("*");
    115.       if(st != temp){
    116.           printf("%d\n",year[j]);
    117.       }
    118. }
    119.  
    120.  
    121.  return 0;
    122. }