The document contains solutions to 50 programming problems in C language. For each problem, the code includes the problem number, code snippet with comments, and input/output. The solutions cover topics like loops, conditional statements, arrays, functions etc. and calculate sums, averages, find even/odd numbers within ranges. Overall the document provides coding solutions and explanations to 50 different programming questions.

1. Rajshahi University of Engineering and Technology
“Heavens Light Is Our Guide”
Department of Electrical & Computer Engineering (ECE)
Assignment
Course No: ECE-1103.
Course Title: Computer Programming.
Assignment Name: Solving 50 programming problems.
Submitted To: Submitted By:
Sagor Chandro Bakchy.
Assistant Professor,
Department of Electrical & Computer
Engineering, RUET
MD. Samiul Haque Biswas.
Roll: 2010052.
Session: 2020-21

2. BeeCrowd ID link: https://www.beecrowd.com.br/judge/en/profile/692469
Problem No: 1051
Solution Code:
#include <stdio.h>
int main()
{
float salary;
scanf("%f", &salary);
if (salary <= 2000)
{
printf("Isenton");
}
else if (salary <= 3000)
{
printf("R$ %.2fn", (salary-2000)*0.08);
}
else if (salary <= 4500)
{
printf("R$ %.2fn", 80+((salary-3000)*0.18));
}
else if(salary>4500)
{
printf("R$ %.2fn", 350+((salary-4500)*0.28));
}
return 0;
}

3. Problem No: 1052
Solution Code:
#include <stdio.h>
int main()
{
int a;
scanf("%d", &a);
switch (a)
{
case 1:
printf("Januaryn");
break;
case 2:
printf("Februaryn");
break;
case 3:
printf("Marchn");
break;
case 4:
printf("Apriln");
break;
case 5:
printf("Mayn");
break;
case 6:
printf("Junen");
break;
case 7:
printf("Julyn");

4. break;
case 8:
printf("Augustn");
break;
case 9:
printf("Septembern");
break;
case 10:
printf("Octobern");
break;
case 11:
printf("Novembern");
break;
case 12:
printf("Decembern");
break;
}
return 0;
}
Problem NO: 1059
Solution Code:
#include <stdio.h>
int main()
{
for (int i = 2; i < 101; i = i + 2)
{
printf("%dn", i);
}
return 0;

5. }
Problem No: 1061
Solution Code:
#include <stdio.h>
int main()
{
int h, hh, hr, m, mm, d, dm, s, ss;
scanf("Dia %d", &d);
scanf("%d : %d : %dn", &h, &m, &s);
scanf("Dia %d", &dm);
scanf("%d : %d : %d", &hh, &mm, &ss);
s = ss - s;
m = mm - m;
h = hh - h;
d = dm - d;
if (s < 0)
{
s += 60;
m--;
}
if (m < 0)
{
m += 60;
h--;
}
if (h < 0)
{
h += 24;
d--;

6. }
printf("%d dia(s)n", d);
printf("%d hora(s)n", h);
printf("%d minuto(s)n", m);
printf("%d segundo(s)n", s);
return 0;
}
Problem No: 1060
Solution Code:
#include <stdio.h>
int main()
{
int t = 6,s=0;
while (t--)
{
float a;
scanf("%f", &a);
if (a > 0)
s++;
}
printf("%d valores positivosn", s);
return 0;
}
Problem No: 1061
Solution Code:
#include <stdio.h>
int main()
{

7. int h, hh, hr, m, mm, d, dm, s, ss;
scanf("Dia %d", &d);
scanf("%d : %d : %dn", &h, &m, &s);
scanf("Dia %d", &dm);
scanf("%d : %d : %d", &hh, &mm, &ss);
s = ss - s;
m = mm - m;
h = hh - h;
d = dm - d;
if (s < 0)
{
s += 60;
m--;
}
if (m < 0)
{
m += 60;
h--;
}
if (h < 0)
{
h += 24;
d--;
}

8. printf("%d dia(s)n", d);
printf("%d hora(s)n", h);
printf("%d minuto(s)n", m);
printf("%d segundo(s)n", s);
return 0;
}
Problem No: 1064
Solution Code:
#include <stdio.h>
int main()
{
int t = 6,s=0; //SamiulHaque
float x=0;
while (t--)
{
float a;
scanf("%f", &a);
if (a > 0)
{
s++;
x = x + a;
}
}
printf("%d valores positivosn", s);
printf("%.1fn", float(x / s));
return 0;
}

9. Problem No: 1065
Solution Code:
#include <stdio.h>
int main()
{
int t = 5, s = 0;
while (t--)
{
int a;
scanf("%d", &a);
if (a % 2 == 0)
s++;
}
printf("%d valores paresn", s);
return 0;
}
Problem No: 1066
Solution Code:
#include <stdio.h>
int main()
{
int t = 5, e = 0, o = 0, p = 0, n = 0;
while (t--)
{
int a;
scanf("%d", &a);
if (a % 2 == 0 || (a*(-1))%2==0) //SamiulHaque
{

10. e++;
}
else
{
o++;
}
if (a > 0)
{
p++;
}
else if(a<0)
{
n++;
}
}
printf("%d valor(es) par(es)n", e);
printf("%d valor(es) impar(es)n", o);
printf("%d valor(es) positivo(s)n", p);
printf("%d valor(es) negativo(s)n", n);
return 0;
}
Problem No: 1067
Solution Code:
#include<stdio.h>
int main()
{
int t;

11. scanf("%d", &t);
for (int i = 1; i < t + 1; i=i+2)
{
printf("%dn", i);
}
return 0;
}
Problem No: 1070
Solution Code:
#include <stdio.h>
int main()
{
int x;
scanf("%d", &x);
if (x % 2 == 0)
{
for (int i = x+1; i < x + 12; i = i + 2)
{
printf("%dn", i);
}
}
else if (x % 2 != 0)
{
for (int i = x; i < x + 11; i = i + 2)
{
printf("%dn", i);
}
}
return 0;}

12. Problem No: 1071
Solution Code:
#include <stdio.h>
int main()
{
int x, y, s = 0;
scanf("%d %d", &x, &y);
if (x > y)
{
for (int i = y+1; i < x; i++)
{
if (i % 2 != 0)
s = s + i;
}
printf("%dn", s);
}
else if (x < y)
{
for (int i = x+1; i < y; i = i++)
{
if (i % 2 != 0)
s = s + i;
}
printf("%dn", s);
}
else
{
printf("0n");
}

13. }
Problem No: 1072
Solution Code:
#include <stdio.h>
int main()
{
int n,c=0,d=0;
scanf("%d",&n);
for (int i = 0; i < n; i++)
{
int x;
scanf("%d", &x);
if (x >= 10 && x <= 20)
{
c++;
}
else
{
d++; } }
printf("%d inn", c);
printf("%d outn", d);
return 0;
}
Problem No: 1073
Solution Code:
#include <stdio.h>
int main()

14. {
long long int n;
scanf("%lld", &n);
if (n % 2 == 0)
{
for (long long int i = 2; i < n + 1; i=i+2)
{
printf("%lld^2 = %lldn",i,i*i);
}
}
else
{
for(long long int i=2; i<n; i=i+2)
{
printf("%lld^2 = %lldn",i,i*i);
}
}
}
Problem No: 1074
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
while (n--)
{
int x;

15. scanf("%d", &x);
if (x == 0)
{
printf("NULLn");
}
else if (x % 2 == 0 && x > 0)
{
printf("EVEN POSITIVEn");
}
else if (x % 2 == 0 && x < 0)
{
printf("EVEN NEGATIVEn");
}
else if (x % 2 != 0 && x < 0)
{
printf("ODD NEGATIVEn");
}
else if (x % 2 != 0 && x > 0)
{
printf("ODD POSITIVEn");
}
}
return 0;
}
Problem No: 1075
Solution Code:
#include <stdio.h>
int main()
{

16. int n;
scanf("%d", &n);
for (int i = 1; i <= 10000; i++)
{
if (i % n == 2)
printf("%dn",i);
}
return 0;
}
Problem No: 1078
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= 10; i++)
{
printf("%d x %d = %dn", i, n, i * n);
}
return 0;
}
Problem No: 1079
Solution Code:
#include <stdio.h>
int main()
{
int n;

17. scanf("%d", &n);
while (n--)
{
float a, b, c;
scanf("%f %f %f", &a, &b, &c);
printf("%.1fn",(a*2+b*3+c*5)/10);
}
return 0;}
Problem No: 1080
Solution Code:
#include <stdio.h>
int main()
{
int highest = -999;
int a[100];
for (int i = 0; i < 100; i++)
{
scanf("%d", &a[i]);
}
for (int i = 0; i < 100; i++)
{
if (highest < a[i])
highest = a[i];
}
printf("%dn", highest);
for (int i = 0; i < 100; i++)
{
if (a[i] == highest)
printf("%dn", i + 1);

18. }
}
Problem No: 1094
Solution Code:
#include <stdio.h>
int main()
{
int n, p = 0, q = 0, r = 0, t;
float e, f, g;
scanf("%d", &n);
while (n--)
{
int x;
char a;
scanf("%d %c", &x, &a);
if (a == 'C')
{
p = p + x;
}
else if (a == 'R')
{
q = q + x;
}
else if (a == 'S')
{
r = r + x;
}
}
t = p + q + r;

19. e = float((p * 100.00) / t);
f = float((q * 100.00) / t);
g = float((r * 100.00) / t);
printf("Total: %d cobaiasn", t);
printf("Total de coelhos: %dn", p);
printf("Total de ratos: %dn", q);
printf("Total de sapos: %dn", r);
printf("Percentual de coelhos: %.2f %%n", e);
printf("Percentual de ratos: %.2f %%n", f);
printf("Percentual de sapos: %.2f %%n", g);
return 0;
}
Problem No: 1095
Solution Code:
#include <stdio.h>
int main()
{
int sum = 60;
for (int i = 1; i <= 37; i = i+3)
{
printf("I=%d J=%dn", i, sum);
sum = sum - 5;
}
return 0;
}
Problem No: 1096
Solution Code:

20. #include <stdio.h>
int main()
{
for (int i = 1; i <= 9; i = i + 2)
{
printf("I=%d J=7n", i);
printf("I=%d J=6n", i);
printf("I=%d J=5n", i);
}
return 0;
}
Problem No: 1097
Solution Code:
#include <stdio.h>
int main()
{
int a = 7, b = 6, c = 5;
for (int i = 1; i <= 9; i = i + 2)
{
printf("I=%d J=%dn", i, a);
printf("I=%d J=%dn", i, b);
printf("I=%d J=%dn", i, c);
a = a + 2;
b = b + 2;
c = c + 2;
}
return 0;
}

21. Problem No: 1098
Solution Code:
#include <stdio.h>
int main()
{
float i, j, a = 1.2, b = 2.2, c = 3.2;
printf("I=0 J=1n");
printf("I=0 J=2n");
printf("I=0 J=3n");
for (i = 0.2; i < 1; i = i + 0.2)
{
printf("I=%0.1f J=%0.1fn", i, a);//SamiulHaque
printf("I=%0.1f J=%0.1fn", i, b);
printf("I=%0.1f J=%0.1fn", i, c);
a = a + 0.2;
b = b + 0.2;
c = c + 0.2;
}
printf("I=1 J=2n");
printf("I=1 J=3n");
printf("I=1 J=4n");
a=2.2,b=3.2,c=4.2;
for (i = 1.2; i < 2; i = i + 0.2)
{
printf("I=%0.1f J=%0.1fn", i, a);//SamiulHaque
printf("I=%0.1f J=%0.1fn", i, b);
printf("I=%0.1f J=%0.1fn", i, c);

22. a = a + 0.2;
b = b + 0.2;
c = c + 0.2;
}
printf("I=2 J=3n");
printf("I=2 J=4n");
printf("I=2 J=5n");
return 0;
}
Problem No: 1099
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
while (n--)
{
int x, y, sum = 0;
scanf("%d %d", &x, &y);
if (x == y)
{
printf("0n");
}
if (x > y && y % 2 == 0)
{
for (int i = y + 1; i < x; i = i + 2)
{

23. sum = sum + i;
}
printf("%dn", sum);
}
else if (x > y && y % 2 != 0)
{
for (int i = y + 2; i < x; i = i + 2)
{
sum = sum + i;
}
printf("%dn", sum);
}
else if (y > x && x % 2 == 0)
{
for (int i = x + 1; i < y; i = i + 2)
{
sum = sum + i;
}
printf("%dn", sum);
}
else if(y>x && x%2!=0)
{
for (int i = x + 2; i < y; i = i + 2)
{
sum = sum + i;
}
printf("%dn", sum);
}
}

24. return 0;
}
Problem No: 1101
Solution Code:
#include <stdio.h>
int main()
{
int m, n;
for (int i = 0; i >= 0; i++)
{
int sum=0;
scanf("%d %d", &m, &n);
if (m > 0 && n > 0)
{
if (m >= n)
{
for (int j = n; j < m + 1; j++)
{
printf("%d ", j);
sum = sum + j;
}
printf("Sum=%dn", sum);
}
else
{
for (int j = m; j < n + 1; j++)
{
printf("%d ", j);
sum = sum + j;

25. }
printf("Sum=%dn", sum);
}
}
else
{
break;
}
}
return 0;
}
Problem No: 1113
Solution Code:
#include <stdio.h>
int main()
{
for (int i = 0; i >= 0; i++)
{
int x, y, r;
scanf("%d %d", &x, &y);
r = x - y;
if (r == 0)
{
break;
}
else if (r > 0)
{
printf("Decrescenten");
}

26. else if (r < 0)
{
printf("Crescenten");
}
}
return 0;
}
Problem No: 1114
Solution Code:
#include <stdio.h>
int main()
{
for (int i = 0; i >= 0; i++)
{
int n;
scanf("%d", &n);
if (n == 2002)
{
printf("Acesso Permitidon");
break;
}
else
{
printf("Senha Invalidan");
}
}
return 0;
}
Problem No: 1115

27. Solution Code:
#include <stdio.h>
int main()
{
for (int i = 0; i >= 0; i++)
{
int x, y;
scanf("%d %d", &x, &y);
if (x == 0 || y == 0)
{
break;
}
else if (x > 0 && y > 0)
{
printf("primeiron");
}
else if (x > 0 && y < 0)
{
printf("quarton");
}
else if (x < 0 && y < 0)
{
printf("terceiron");
}
else if (x < 0 && y > 0)
{
printf("segundon");
}
}

28. return 0;
}
Problem No: 1116
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
while (n--)
{
float x, y;
scanf("%f %f", &x, &y);
if (y == 0)
{
printf("divisao impossiveln");
}
else
{
printf("%.1fn", (x / y));
}
}
return 0;
}
Problem No: 1117
Solution Code:
#include <stdio.h>
int main()

29. {
float x, sum;
int j = 1;
while (j < 3)
{
scanf("%f", &x);
if (x >= 0.0 && x <= 10.0)
{
sum = sum + x;
j++;
continue;
}
else
{
printf("nota invalidan");
continue;
}
}
printf("media = %.2fn", sum / 2.0);
return 0;
}
Problem No: 1118
Solution Code:
#include <stdio.h>
int main()
{
int x = 1, j = 0;

30. float s[2], a, avg;
while (x != 2)
{
if (x == 1)
{
while (j <= 1)
{
scanf("%f", &a);
if (a >= 0.0 && a <= 10.0)
{
s[j] = a;//SamiulHaque
j++;
}
else
{
printf("nota invalidan");
}
}
avg = (s[0] + s[1]) / 2.0;
printf("media = %.2fn", avg);
}
printf("novo calculo (1-sim 2-nao)n");
scanf("%d", &x);
j = 0;
continue;
}
return 0;

31. }
Problem No: 1132
Solution Code:
#include <stdio.h>
int main()
{
int x, y, add = 0;
scanf("%d %d", &x, &y);
if (x == y)
{
printf("0");
}
else if (x > y)
{
for (int i = y; i <=x; i++)
{
if (i % 13 != 0)
{
add = add + i;
}
}
printf("%dn", add);
}
else if (x < y)
{
for (int i = x; i <=y; i++)
{
if (i % 13 != 0)
{

32. add = add + i;
}
}
printf("%dn", add);
}
return 0;
}
Problem No: 1133
Solution Code:
#include <stdio.h>
int main()
{
int x, y, a = 0, b = 0;
scanf("%d %d", &x, &y);
if (x > y)
{
for (int i = y + 1; i < x; i++)
{
if (i % 5 == 2 || i % 5 == 3)
{
printf("%dn", i);
}
}
}
else if (x < y)
{
for (int i = x + 1; i < y; i++)
{
if (i % 5 == 2 || i % 5 == 3)

33. {
printf("%dn", i);
}
}
}
return 0;}
Problem No: 1134
Solution Code:
#include <stdio.h>
int main()
{
int a = 0, b = 0, c = 0;
for (int i = 1; i > 0; i++)
{
int n;
scanf("%d", &n);
if(n==1)
{
a++;
}
else if(n==2)
{
b++;
}
else if(n==3)
{
c++;
}
else if(n==4)

34. {
break;
}
else
{
continue;
}
}
printf("MUITO OBRIGADOn");
printf("Alcool: %dn", a);
printf("Gasolina: %dn", b);
printf("Diesel: %dn", c);
return 0;
}
Problem No: 1142
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
int a = 1, b = 2, c = 3;
while (n--)
{
printf("%d ", a);
printf("%d ", b);
printf("%d ", c);
printf("PUMn");
a = a + 4;

35. b = b + 4;
c = c + 4;
}
return 0;
}
Problem No: 1143
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
printf("%d %d %dn", i, i * i, i * i * i);
}
return 0;
}
Problem No:1144
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{

36. printf("%d %d %dn", i, i * i, i * i * i);
printf("%d %d %dn", i, (i * i) + 1, (i * i * i) + 1);
}
return 0;
}
Problem No: 1145
Solution Code:
#include <stdio.h>
int main()
{
int x, y, c = 0;
scanf("%d %d", &x, &y);
for (int i = 1; i <= y; i++)
{
if (i % x != 0)
{
printf("%d ", i);//SamiulHaque
}
else if (i % x == 0)
{
printf("%dn",i);
}
}
return 0;
}
Problem No: 1146
Solution Code:

37. #include <stdio.h>
int main()
{
int x;
while (x != 0)
{
scanf("%d", &x);
for (int i = 1; i <= x; i++)
{
if (i % x != 0)
{
printf("%d ", i);//SamiulHaque
}
else if (i % x == 0)
{
printf("%dn", i);
}
}
}
return 0;
}
Problem No:1149
Solution Code:
#include <stdio.h>
int main()
{
int n, a, sum = 0;

38. scanf("%d ", &a);
for (int i = 0; i >= 0; i++)
{
scanf("%d", &n);
if (n > 0)
{
for (int i = a; i < a + n; i++)
{
sum = sum + i;
}
printf("%dn", sum);
break;
}
else
{
continue;
}
}
return 0;
}
Problem No: 1150
Solution Code:
#include <stdio.h>
int main()
{
int x, z, sum = 0, c=0;

39. scanf("%d", &x);
for (int i = 0; i >= 0; i++)
{
scanf("%d", &z);
if (z <= x)
{
continue;
}
else
{
break;
}
}
for (int i = 0; i >= 0; i++)
{
sum = x + i + sum;
c++;
if (sum >= z)
{
break;
}
}
printf("%dn", c);
return 0;
}
Problem No: 1151
Solution Code:

40. #include <stdio.h>
int main()
{
int n, a = 0, b = 1, c;
scanf("%d", &n);
if (n == 1)
{
printf("%n", a);
}
else if (n == 2)
{
printf("%d %dn", a, b);
}
else
{
printf("%d %d", a, b);
for (int i = 2; i < n; ++i)
{
c = a + b;
printf(" %d", c);
a = b;
b = c;
}
printf("n");
}
return 0;
}

41. Problem No: 1153
Solution Code:
#include <stdio.h>
int main()
{
int n, fact = 1;
scanf("%d", &n);
if (n <= 1)
{
printf("1n");
}
else
{
for (int i = 1; i <= n; i++)
{
fact = fact * i;
}
printf("%dn", fact);
}
return 0;
}
Problem No: 1154
Solution Code:
#include <stdio.h>
int main()
{
float sum = 0, a = 0;

42. for (int i = 0; i >= 0; i++)
{
int x;
scanf("%d", &x);
if (x < 0)
{
break;
}
else if (x >= 0)
{
sum = sum + x;
a++;
}
}
printf("%.2fn", float(sum / a));
return 0;
}
Problem No: 1155
Solution Code:
#include <stdio.h>
int main()
{
float sum = 0;
for (float i = 1; i <= 100; i++)
{
sum = sum + (1 / i);
}

43. printf("%.2fn", sum);
return 0;
}
Problem No: 1156
Solution Code:
#include <stdio.h>
#include <math.h>
int main()
{
float sum = 0;
for (float i = 1, j = 0; i <= 39, j <= 20; i = i + 2, j++)
{
sum = sum + (i / pow(2, j));
}
printf("%.2fn", sum);
return 0;
}
Problem No: 1157
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
if (n % i == 0)

44. {
printf("%dn", i);
}
}
return 0;
}
Problem No: 1158
Solution Code:
#include <stdio.h>
int main()
{
int n;
scanf("%d",&n);
while (n--)
{
int x, y, sum = 0;
scanf("%d %d", &x, &y);
if (x % 2 == 0)
{
for (int i = x; i < x + 2 * y; i++)
{
if (i % 2 != 0)
{
sum = sum + i;
}
}
printf("%dn", sum);

45. }
else if (x % 2 != 0)
{
for (int i = x; i < x + 2 * y; i++)
{
if (i % 2 != 0)
{
sum = sum + i;
}
}
printf("%dn", sum);
}
}
return 0;
}
Problem No: 1159
Solution Code:
#include <stdio.h>
int main()
{
for (int i = 0; i >= 0; i++)
{
int x, sum = 0;
scanf("%d", &x);
if (x == 0)
{
break;

46. }
else
{
for (int i = x; i < x + 10; i++)
{
if (i % 2 == 0)
{
sum = sum + i;
}
else
{
continue;
}
}
printf("%dn", sum);
}
}
return 0;
}
Problem No: 1160
Solution Code:
#include <stdio.h>
int main()
{
int t, pa, pb, c;
double ga, gb, p, q;
scanf("%d", &t);

47. while (t--)
{
c = 0;
scanf("%d %d %lf %lf", &pa, &pb, &ga, &gb);
p = ga / 100;
q = gb / 100;
while (pa <= pb)
{
pa = pa + (pa * p);//SamiulHaque
pb = pb + (pb * q);
c++;
if (c > 100)
{
printf("Mais de 1 seculo.n");
break;
}
}
if (c <= 100)
{
printf("%d anos.n", c);
}
}
return 0;}

48. Submission Record: