Here is simple C program to find the value of a collection of coins is given,which takes number of number of quarters, dimes, nickels, and pennies from user and compute the value of collection of coins manually.
C program to find the value of a collection of coins
#include <stdio.h> void main () { // Local data ... int pennies; // input: count of pennies int nickels; // input: count of nickels int dimes; // input: count of dimes int quarters; // input: count of quarters int temp, left; // temporaries for various // computations // Read in the count of quarters, dimes, nickels and pennies. printf("Enter the number of quarters, dimes, nickels, and pennies: "); scanf("%d %d %d %d", &quarters, &dimes, &nickels, &pennies); // Compute the total value in cents. left = 25 * quarters + 10 * dimes + 5 * nickels + pennies; // Find and display the value in dollars printf("Your collection is worth\n "); temp = left / 100; printf("\t%d dollar", temp); if (temp==1) printf(", "); else printf("s, "); left = left % 100; // Find and display the value left in quarters temp = left / 25; printf("%d quarter", temp); if (temp==1) printf(", "); else printf("s, "); left = left % 25; // Find and display the value left in dimes temp = left / 10; printf("%d dime", temp); // Here, just for fun, instead of using a conditional statement, // I use a conditional expression and string concatenation printf ((temp==1) ? ", " : "s, "); left = left % 10; // Find and display the value left in nickels temp = left / 5; printf("%d nickel", temp); if (temp==1) printf(", and "); else printf("s, and "); left = left % 5; // Find and display the value left in pennies printf("%d penn", left); if (left==1) printf("y\n"); else printf("ies\n"); }