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Authored by Irina Snape
Instructional Technology
University

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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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