Contoh penulisan proses LOOP
- Statement While
- Statement Do…….While
- Statement FOR
- Statement FOR Positif
- Statement FOR Negatif
- Statement FOR berasarang (Nested Loop)
1. Statement While
int bil=1;
while(bil<=5)
cout<
++bil; bil = bil + 1
Output: 12345
Jwb :
| Bil = 1 | Bil<=5 | Cetak bil | Bil= bil + 1 |
| 1 | 1<=5 | 1 | 2 |
| 2 | 2<=5 | 2 | 3 |
| 3 | 3<=5 | 3 | 4 |
| 4 | 4<=5 | 4 | 5 |
| 5 | 5<=5 | 5 | 6 |
| 6 | 6<=5 | | |
| | | | |
2. Statement Do….While
int bil=2;
do
cout< bil+=2;
while (bil>=10);
output: 2 4 6 8 10
jwb:
| Bil=2 | Cetak bil | Bil + 2 | Bil>=10 | Y/T |
| 2 | 2 | 4 | 4>=10 | T |
| 4 | 4 | 6 | 6>=10 | T |
| 6 | 6 | 8 | 8>=10 | T |
| 8 | 8 | 10 | 10>=10 | T |
| 10 | 10 | 12 | 12>=10 | Y |
3. Statement FOR
Contoh 1.a
#include
Void main()
{
Int I,N;
N = 45;
I = 1;
While (I<=5);
{
Printf (“\n%i,”N);
N = N + 5
I = I + 1
}
}
Output: 45 50 55 60 65
Jwb:
| N = 45 | I = 1 | I<=5 | Y/T | Cetak N | N = N + 5 | I = I + 1 |
| 45 | 1 | 1<=5 | Y | 45 | 50 | 2 |
| 50 | 2 | 2<=5 | Y | 50 | 55 | 3 |
| 55 | 3 | 3<=5 | Y | 55 | 60 | 4 |
| 60 | 4 | 4<=5 | Y | 60 | 65 | 5 |
| 65 | 5 | 5<=5 | Y | 65 | 70 | 6 |
| 70 | 6 | 6<=5 | T | | | |
Contoh 1.b
#include
Void main()
{
Int I,N;
N = 45;
I = 1;
While (I<=5);
{
N = N + 5
I = I + 1
}
Printf (“\n%i,”N);
}
Output: 70
Jwb :
| N = 45 | I = 1 | I<=5 | Y/T | N = N + 5 | I = I + 1 | Cetak N |
| 45 | 1 | 1<=5 | Y | 50 | 2 | |
| 50 | 2 | 2<=5 | Y | 55 | 3 | |
| 55 | 3 | 3<=5 | Y | 60 | 4 | |
| 60 | 4 | 4<=5 | Y | 65 | 5 | |
| 65 | 5 | 5<=5 | Y | 70 | 6 | |
| 70 | 6 | 6<=5 | T | | | 70 |
Contoh 1.c
#include
Void main()
{
Int I,N;
N = 45;
I = 1;
While (I<=5);
{
N = N + 5
Printf (“\n%i,”N);
I = I + 1
}
}
Output: 45 50 55 60 65
Jwb:
| N = 45 | I = 1 | I<=5 | Y/T | N = N + 5 | Cetak N | I = I + 1 |
| 45 | 1 | 1<=5 | Y | 50 | 45 | 2 |
| 50 | 2 | 2<=5 | Y | 55 | 50 | 3 |
| 55 | 3 | 3<=5 | Y | 60 | 55 | 4 |
| 60 | 4 | 4<=5 | Y | 65 | 60 | 5 |
| 65 | 5 | 5<=5 | Y | 70 | 65 | 6 |
| 70 | 6 | 6<=5 | T | | | |
Contoh 2
#include
Void main()
{
Int I,N;
N = 10;
I = 10;
While (N<=100);
{
N = N + I
Printf (“\n%i,”N);
I = I + 5;
}
}
Output: 20 35 55 80 110
Jwb:
| N = 10 | I = 1 | N<=100 | Y/T | N = N + 1 | Cetak N | I = I + 5 |
| 10 | 10 | 10<=100 | Y | 20 | 20 | 15 |
| 20 | 15 | 15<=100 | Y | 35 | 35 | 20 |
| 35 | 20 | 20<=100 | Y | 55 | 55 | 25 |
| 55 | 25 | 25<=100 | Y | 80 | 80 | 30 |
| 80 | 30 | 30<=100 | Y | 110 | 110 | 35 |
| 110 | 35 | 35<=100 | T | | | |
Contoh 3.a
#include
Void main()
{
Int I,N;
N = 10;
X = 10;
T = 0;
While (T<=100);
{
T = T + N;
N = N + X;
X = X + 5;
}
Printf (“\n%i,”T);
}
Output: 120
Jwb:
| N=10 | X=10 | T=0 | T<=100 | Y/T | T =T+N | N =N+X | X=X+5 | Cetak T |
| 10 | 10 | 0 | 0<=100 | Y | 10 | 20 | 15 | |
| 20 | 15 | 10 | 10<=100 | Y | 30 | 35 | 20 | |
| 35 | 20 | 30 | 30<=100 | Y | 65 | 55 | 25 | |
| 55 | 25 | 65 | 65<=100 | Y | 120 | 80 | 30 | |
| 80 | 30 | 120 | 120<=100 | T | | | | 120 |
Contoh 3.b
#include
Void main()
{
Int I,N;
N = 10;
X = 10;
T = 0;
While (T<=100);
{
T = T + N;
Printf (“\n%i,”T);
N = N + X;
X = X + 5;
}
}
Output: 10 30 65 120
Jwb:
| N=10 | X=10 | T=0 | T<=100 | Y/T | T =T+N | Cetak T | N =N+X | X=X+5 |
| 10 | 10 | 0 | 0<=100 | Y | 10 | 10 | 20 | 15 |
| 20 | 15 | 10 | 10<=100 | Y | 30 | 30 | 35 | 20 |
| 35 | 20 | 30 | 30<=100 | Y | 65 | 65 | 55 | 25 |
| 55 | 25 | 65 | 65<=100 | Y | 120 | 120 | 80 | 30 |
| 80 | 30 | 120 | 120<=100 | T | | | | |
Nested FOR
Contoh 1.a
int I.J;
for (I = 1, 1<=3;I++);
{
for(J=1;J<=5;J++);
{
Printf(“\n%i”,I);
}
}
Output: 1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
Jwb:
| I=1 | I<=3 | Y\T | J=1 | J<=5 | Y\T | Cetak I | J++ | I++ |
| 1 | 1<=3 | Y | 1 | 1<=5 | Y | 1 | 2 | |
| | | | 2 | 2<=5 | Y | 1 | 3 | |
| | | | 3 | 3<=5 | Y | 1 | 4 | |
| | | | 4 | 4<=5 | Y | 1 | 5 | |
| | | | 5 | 5<=5 | Y | 1 | 6 | |
| | | | 6 | 6<=5 | T | | | 2 |
| 2 | 2<=3 | Y | 1 | 1<=5 | Y | 2 | 2 | |
| | | | 2 | 2<=5 | Y | 2 | 3 | |
| | | | 3 | 3<=5 | Y | 2 | 4 | |
| | | | 4 | 4<=5 | Y | 2 | 5 | |
| | | | 5 | 5<=5 | Y | 2 | 6 | |
| | | | 6 | 6<=5 | T | | | 3 |
| 3 | 3<=3 | Y | 1 | 1<=5 | Y | 3 | 2 | |
| | | | 2 | 2<=5 | Y | 3 | 3 | |
| | | | 3 | 3<=5 | Y | 3 | 4 | |
| | | | 4 | 4<=5 | Y | 3 | 5 | |
| | | | 5 | 5<=5 | Y | 3 | 6 | |
| | | | 6 | 6<=5 | T | | | 4 |
| 4 | 4<=3 | T | | | | | | |
Contoh 1.b
int I.J;
for (I = 1, 1<=3;I++);
{
for(J=1;J<=5;J++);
{
Printf(“\n%i”,J);
}
}
Output: 1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
Jwb:
| I=1 | I<=3 | Y\T | J=1 | J<=5 | Y\T | Cetak J | I++ | J++ |
| 1 | 1<=3 | Y | 1 | 1<=5 | Y | 1 | 2 | |
| 2 | 2<=3 | Y | | | | 1 | 3 | |
| 3 | 3<=3 | Y | | | | 1 | 4 | |
| 4 | 4<=3 | T | | | | | | 2 |
| 1 | 1<=3 | Y | 2 | 2<=5 | Y | 2 | 2 | |
| 2 | 2<=3 | Y | | | | 2 | 3 | |
| 3 | 3<=3 | Y | | | | 2 | 4 | |
| 4 | 4<=3 | T | | | | | | 3 |
| 1 | 1<=3 | Y | 3 | 3<=5 | Y | 3 | 2 | |
| 2 | 2<=3 | Y | | | | 3 | 3 | |
| 3 | 3<=3 | Y | | | | 3 | 4 | |
| 4 | 4<=3 | T | | | | | | 4 |
| 1 | 1<=3 | Y | 4 | 4<=5 | Y | 4 | 2 | |
| 2 | 2<=3 | Y | | | | 4 | 3 | |
| 3 | 3<=3 | Y | | | | 4 | 4 | |
| 4 | 4<=3 | T | | | | | | 5 |
| 1 | 1<=3 | Y | 5 | 5<=5 | Y | 5 | 2 | |
| 2 | 2<=3 | Y | | | | 5 | 3 | |
| 3 | 3<=3 | Y | | | | 5 | 4 | |
| 4 | 4<=3 | T | | | | | | 6 |
| | | | 6 | 6<=5 | T | | | |
| | | | | | | | | | |
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