| At line 1 added 28 lines |
| !!Броеви |
| !C++ |
| %%prettify |
| {{{ |
| //v10 |
| #include <iostream> |
| |
| using namespace std; |
| |
| int main() |
| { |
| int z, r, x, y; |
| |
| cin >> z >> r; |
| |
| x=(z+r)/2; |
| y=z-x; |
| |
| if(x>=0 && y>=0 && (z+r)%2==0) |
| cout << x << " " << y << endl; |
| else |
| cout << "NE MOZE" << endl; |
| |
| return 0; |
| } |
| }}} |
| /% |
|
| At line 34 added one line |
| //v10 |
| At line 36 added one line |
| |
| At line 38 added one line |
| |
| At line 10 changed 4 lines |
| int p,v; |
| cin>>p>>v; |
| int kusur=v-p, i=0; |
| while(kusur>0) |
| int x,y,k,petka=0,dvojka=0,kec=0; |
| |
| cin>>x>>y; |
| k=y-x; |
| while(k>0){ |
| while(k>=5) |
| { |
| k=k-5; |
| petka++; |
| } |
| while(k>=2) |
| At line 15 changed 15 lines |
| if(kusur>=5) |
| { |
| kusur-=5; |
| i++; |
| } |
| else if(kusur>=2) |
| { |
| kusur-=2; |
| i++; |
| } |
| else if(kusur>=1) |
| { |
| kusur-=1; |
| i++; |
| } |
| k=k-2; |
| dvojka++; |
| At line 31 changed one line |
| cout<<i; |
| while(k>=1) |
| { |
| k=k-1; |
| kec++; |
| } |
| } |
| cout<<petka+dvojka+kec; |
| |
| At line 37 changed one line |
| !!Броеви |
| !!Бројот 9 |
| At line 73 added one line |
| //v10 |
| At line 75 added one line |
| |
| At line 77 added one line |
| |
| At line 45 changed 10 lines |
| int z,r; |
| cin>>z>>r; |
| int x,y; |
| y=(z+r)/2; |
| x=z-y; |
| int p1,p2; |
| if(x>y) |
| { |
| p1=x; |
| p2=y; |
| int n, m, a=0, b, c; |
| long long sum=0; |
| string broj; |
| |
| cin >> str; |
| |
| for(int i=0;i<broj.size();i++){ |
| suma+=broj[i]-48; //-48 или -'0' за да се конвертира од ASCII во декаден |
| At line 56 changed 13 lines |
| else |
| { |
| p1=y; |
| p2=x; |
| } |
| if(y<0 || x<0 || r!=p1-p2 || z!=p1+p2) |
| { |
| cout<<"NE MOZE"; |
| } |
| else |
| { |
| cout<<p1<<" "<<p2; |
| } |
| |
| if(suma%9==0) |
| cout << 0 << endl; |
| else if(suma%3==0) |
| cout << 2 << endl; //Ако е делив со 3, а не со 9 треба да се додаде збирот уште 2 пати |
| else |
| cout << 8 << endl; //Ако не е делив со 3 и 9, тогаш бројот е заемно прост со 9, |
| //треба да се собере уште 8 пати |
| |
| At line 102 added 63 lines |
|
| !!Мечето ушко |
| !C++ |
| %%prettify |
| {{{ |
| //Филип Бујаровски |
| #include <iostream> |
| #include <cstring> |
| #include <cmath> |
| #include <math.h> |
|
| using namespace std; |
| |
| int main(){ |
|
| int i, j, n, k, zbir=0; |
| cin>>n; |
| |
| int niza[n]; |
| |
| for(i=0;i<n;i++) |
| { |
| cin>>niza[i]; |
| zbir+=niza[i]; |
| } |
| |
| int x; |
| bool c[(zbir)+1]; |
| |
| memset(c,false,sizeof(c)); |
| c[0]=true; |
| |
| for(i=0;i<n;i++) |
| { |
| k=niza[i]; |
| for(j=zbir;j>=0;j--) |
| { |
| x=j+k; |
| if(c[j]==true) |
| { |
| if(j+k<zbir) |
| c[j+k]=true; |
| } |
| } |
| } |
| |
| int p=0,q=0; |
| for(i=zbir/2;i>=0;i--) |
| { |
| if(c[i]==true) |
| { |
| p=i; |
| break; |
| } |
| } |
| q=zbir-p; |
| cout<<fabs(p-q)><<endl; |
| |
| |
| return 0; |
| } |
| }}} |
| /% |