Yes because even if you didn't get an A you tried you best.
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int divide1(int a,int b) { int t=1; while(b*t<=a) { t++; } return t-1; }
The set A union B can be decomposed into three disjoint sub sets A\ (A int B), B\(A int B), and (A int B). So in this case (A union B) and (A int B) are fixed but "moving" elelments from A\ (A int B) into B\(A int B) will not affect (A union B) and (A int B). You should be able to fill in the details now.
#include <stdlib.h> #include <stdio.h> #include <conio.h> int sum(int, int); int sub(int, int); int mul(int, int); int div1(int, int); int main() { int a,b,d; int ch; printf("\t\tCalculator Implementation\n"); printf("\t\t~\n"); printf("\t\t 1: Add\n"); printf("\t\t 2: Sub\n"); printf("\t\t 3: Multiply\n"); printf("\t\t 4: Divide\n"); printf("\t\t 5: Exit\n\n\n"); printf("\t\t Enter the choice: "); scanf("%d", &ch); if(ch==5) exit(0); printf("\nEnter first number: "); scanf("%d", &a); printf("\nEnter Second number: "); scanf("%d", &b); printf("\na = %d; b = %d;\n", a, b); printf("==============\n\n"); switch(ch) { case 1 : d=sum(a,b); printf("Sum %d + %d = %d",a,b,d); break; case 2: d=sub(a,b); printf("Subtraction %d - %d = %d",a,b,d); break; case 3: d=mul(a,b); printf("Multiplication %d * %d = %d",a,b,d); break; case 4 : d=div1(a,b); printf("Division %d / %d = %d",a,b,d); break; default: printf("Invalid Choice!"); getch(); } getch(); return 0; } int sum(int a, int b) { return a + b; } int sub(int a, int b) { return a - b; } int mul(int a, int b) { return a * b; } int div1(int a, int b) { return a / b; }
#include<stdio.h> int main() { int a,b,c,d; for(a=1; a<5; a++) { for(b=1; b<5; b++) { for(c=1; c<5; c++) { for(d=1; d<5; d++) { if(!(a==b a==c a==d b==c b==d c==d)) printf("dd\n",a,b,c,d); } } } } return 0; }