.programmingenhance

#include<stdio.h> #include<conio.h> #include<math.h> int isdatevalid( int month, int day, int year) { if (day <= 0) return 0 ; switch ( month ) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: if (day > 31) return 0 ; else return 1 ; case 4: case 6: case 9: case 11: if (day > 30) return 0 ; else return 1 ; case 2: if ( day > 29 ) return 0 ; if ( day < 29 ) return 1 ; else return 0 ; } return 0 ; } //------------------------------------------------ int fm( int date, int month, int year) { int fmonth,leap; //leap function 1 for leap & 0 for non-leap if ((year%100==0) && (year%400!=0)) leap=0; else if (year%4==0) leap=1; else leap=0; fmonth=3+(2-leap)*((month+2)/(2*month))+(5*month+month/9)/2; //f(m) formula fmonth = fmonth % 7; //bring it in range of 0 to 6 return fmonth; } //---------------------------------------------- int day_of_week( int date, int month, int year) { int

An electric power distribution company charges its domestic consumers as follows. Consumption Rate of Units Charge ------------------------------------------------------ 0-200 Rs.0.50 per unit 201-400 Rs.100 plus Rs.0.65 per unit excess 200 401-600 Rs.230 plus Rs.0.80 per unit excess of 400. #include<stdio.h> #include<conio.h> void main() { int n, p; float amount; clrscr(); printf( "Enter the customer number: " ); scanf( "%d" ,&n); printf( "Enter the power consumed: " ); scanf( "%d" ,&p); if (p>=0 && p<=200) amount=p*0.50; else if (p>200 && p<400) amount = 100+((p-200) * 0.65); else if (p>400 && p<=600) amount = 230 + ((p-400) * 0.80); printf( "Amount to be paid by customer no. %d is Rs.:%5.2f." ,n,amount); getch(); } Output : Enter the customer number: 1 Enter the power consumed: 100 Amount to be paid by customer no. 1 is Rs.:50.00.