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Instructional Technology

University

Hard

Created by

Irina Snape

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28 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Triunghiul de sub diagonala secundară (inclusiv diagonala) unei matrice pătrate se poate parcurge numai cu secvenţele:

1. for(i=0; i<n; i++)   for(j=n-i-1; j<n; j++) 

2. for(i=0; i<n; i++)   for(j=n-1; j>=n-i-1; j--)

3. for(i=n-1; i>=0; i--)   for(j=n-i-1; j<n; j++)

4. for(i=n-1; i>=0; i--)   for(j=n-1; j>=n-i-1; j--)

5. for(j=0; j<n; j++)   for(i=n-j-1; i<n; i++)

6. for(j=0; j<n; j++)   for(i=n-1; i>=n-j-1; i--)

7. for(j=n-1; j>=0; j--)   for(i=n-j-1; i<n; i++)

8. for(j=n-1; j>=0; j--)   for(i=n-1; i>=n-j-1; i--)

1,2,5,6

toate

1,2,3,4

3,4,7,8

5,6,7,8

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Triunghiul de sub diagonala secundară (inclusiv diagonala) unei matrice pătrate se poate parcurge numai cu secvenţele:

1. for(i=0; i<n; i++)   for(j=n-i-1; j<n; j++)

2. for(i=0; i<n; i++)   for(j=n-1; j>=n-i-1; j--)

3. for(i=n; i>=0; i--)   for(j=n-i-1; j<n; j++)

4. for(i=n-1; i>=0; i--)   for(j=n; j>=n-i-1; j--) 

5. for(j=0; j<n; j++)   for(i=n-j-1; i<n; i++)

6. for(j=0; j<n; j++)   for(i=n-1; i>=n-j-1; i--)

7. for(j=n-1; j>=0; j--)   for(i=n-j; i<n; i++)

8. for(j=n-1; j>=0; j--)   for(i=n-1; i>=n-j; i--)

1,2,3,4

1,2,5,6

5,6,7,8

toate

3,4,7,8

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Fie secventa de program C: for (i=0; i<n; i++) y[i]=x[i]. Dintre urmatoarele secvente C:

1. for (i=0; i<n) { y[i]=x[i]; i=i+1; }

2. for (i=0; i<n) { i++; y[i]=x[i]; }

3. for (i=n; i>0) { i--; y[i]=x[i]; }

4. i=0; while (i<n) { y[i]=x[i]; i=i+1; }

5. i=0; do { y[i]=x[i]; i=i+1; } while (i<n) sunt echivalente cu secventa initiala doar:

1,3,4

3,4,5

2,3,4,5

1,2,3

1,3,4,5

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Secventa de program C: while(i<n && x[i]!=a) i++; este echivalenta cu:

do i++; while(i<n && x[i]!=a)

if (i<n && x[i]!=a) while (i>=n ||x[i]==a) i++;

if(i<n && x[i]!=a) do i++; while (i<n && x[i]!=a)

i++; while (i>=n || x[i]==a) i++;

if(i>=n || x[i]==a) while(i<n && x[i]!=a) i++;

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Care dintre urmatoarele secvente de program C realizeaza ordonarea crescatoare a elementelor unui vector de dimensiune n:

1. for (i=0; i<n-1; i++) for(j=i+1; j<n; j++) if (x[i]>x[j]) {a=x[i]; x[i]=x[j]; x[j]=a;}

2. do {vb=0; for(i=0; i<n-1; i++) if(x[i]>x[i+1]) {vb=1; a=x[i]; x[i]=x[i+1]; x[i+1]=a;} while (vb)

3. for (i=0; i<n-1; i++) {m=x[i]; p=i; for(j=i+1; j<n; j++) if(x[j]<m) {m=x[j]; p=j;} a=x[i]; x[i]=x[p]; x[p]=a;}

4. {vb=1; while(vb) for(i=0; i<n-1; i++) if(x[i]>x[i+1]) {a=x[i]; x[i]=x[i+1]; x[i+1]=a; vb=1; }

5. for (i=n-1; i>0; i--) for(j=i-1; j>=0; j--) if (x[i]<x[j]) {a=x[i]; x[i]=x[j]; x[j]=a;}

1,2,3,4 si 5

1,2,3 si 4

1,2 si 4

1,2,3 si 5

1,2 si 3

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Secventa de program C: int s=0;  for(v=vi; v<=vf; v=v+vr) s++; este echivalenta cu:

int s=0; v=vf; while(v>vi) {s++; v=v-vr;}

int s=0; v=vi; while(v<vf) {s++; v=v+vr;}

int s=0; v=vi; do {s++; v=v+vr;} while(v<vf)

int s=0; v=vi; if(v<=vf) do {s++; v=v+vr;} while(v<=vf)

int s=0; v=vf; do {s++; v=v-vr;} while(v>vi)

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Fie o matrice A(m,m). Să se stabilească ce calculează secvența următoare:

p=a[0][1]; for(i=0; i<m/2; i++) for (j=i; j<m-i; j++) if (a[i][j]>p) p=a[i][j];

Alegeți o opțiune:

maximul din triunghiul de deasupra diagonalei principale (exclusiv diagonala);

maximul din triunghiul de sub diagonala secundara (exclusiv diagonala);

maximul din triunghiul superior format de cele doua diagonale (inclusiv diagonalele);

. maximul din triunghiul superior format de cele doua diagonale (exclusiv diagonalele);

maximul din triunghiul inferior format de cele doua diagonale (inclusiv diagonala);

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