List three basic blocks in the digital clock project. ?->(Show Answer!)

1. List three basic blocks in the digital clock project.

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MCQ-> There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular-skilled employees (RE). During the next five months, the division has to complete five projects every month. Out of the 25 projects, 5 projects are "challenging", while the remaining ones are "standard". Each of the challenging projects has to be completed in different months. Every month, five teams — T1 T2, T3, T4 and T5, work on one project each. T1, T2, T3, T4 and T5 are allotted the challenging project in the first, second, third, fourth and fifth month, respectively. The team assigned the challenging project has one more employee than the rest. In the first month, T1 has one more SE than T2, T2 has one more SE than T3, T 3 has one more SE than T4, and T4 has one more SE than T5. Between two successive months, the composition of the teams changes as follows: a. The team allotted the challenging project, gets two SE from the team which was allotted the challenging project in the previous month. In exchange, one RE is shifted from the former team to the latter team. b. After the above exchange, if T1 has any SE and T5 has any RE, then one SE is shifted from T1 to T5, and one RE is shifted from T5 to T1. Also, if T2 has any SE and T4 has any RE, then one SE is shifted from T2 to T4, and one RE is shifted from T4 to T2. Each standard project has a total of 100 credit points, while each challenging project has 200 credit points. The credit points are equally shared between the employees included in that team.The number of times in which the composition of team T2 and the number of times in which composition of team T4 remained unchanged in two successive months are:
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MCQ->What will be the output of the program? #include<stdio.h> #include<stdarg.h> void fun1(char, int, int , float , char ); void fun2(char ch, ...); void (p1)(char, int, int , float , char ); void (p2)(char ch, ...); int main() { char ch='A'; int i=10; float f=3.14; char p="Hello"; p1=fun1; p2=fun2; (p1)(ch, i, &i, &f, p); (p2)(ch, i, &i, &f, p); return 0; } void fun1(char ch, int i, int pi, float pf, char p) { printf("%c %d %d %f %s \n", ch, i, pi, pf, p); } void fun2(char ch, ...) { int i, pi; float pf; char p; va_list list; printf("%c ", ch); va_start(list, ch); i = va_arg(list, int); printf("%d ", i); pi = va_arg(list, int); printf("%d ", pi); pf = va_arg(list, float); printf("%f ", pf); p = va_arg(list, char ); printf("%s", p); }....

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