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Created March 14, 2026 09:39
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divide-et-impera.cpp
#include <iostream>
using namespace std;
int v[100];
// 1 5
/*
n=5
10 101 90 23 44
1 2 3 5
1 5
return v[1]
return v[2]
return v[3]
return v[4]
return v[5]
*/
int max(int i, int j) {
int a, b;
if(i == j) return v[ i ];
else {
a = max(i, (i + j) / 2);//max(1,2)//max(1,1)
b = max( (i + j) / 2 + 1, j); //max(3,5)//max(2,2)
if(a > b) return a;
else
return b;
}
}
/*
n=5
10 101 90 23 44
[ind1,ind2,ind3,ind4,ind5]
limita inferioara = 1
limita superioara = 5
[1,5]
[ 1, 2 ] [ 3, 5 ]
[1,1] [2,2] [3,4] [4,5]
v[1] v[2] [3,3] [4,4] [4,4][5,5]
v[3] v[4] v[4] v[5]
max(m0,m1) max(m3,m4) max(m4,m5)
*/
int main() {
int n;
cout<<"n=";
cin>>n;
for(int i = 1; i <= n; ++i) cin>>v[ i ];
cout<<"MAX = "<<max( 1, n );
return 0;
}
/*
daca i == i valoarea maxima va fi v[i]
contrar vom imparti vectorul in doi vectori (primul vector va contine componentele de la i
la (i + j) div 2), al doilea vector va contine componentele de la (i+j)div 2 + 1, pana la j,
rezolvam subproblemele (aflam maximul pentru fiecare din ele) iar solutia va fi data de valoarea maxima dintre rezultatele celor doua subprobleme.
problema
subprob1 subproblema2
subprobl1,1 subprobl2,1
int factorial(int n) {
if(n == 1) return 1
else return n * factorial(n-1)
}
factorial(5)
n! = 1 * 2 * ... * n;
5! = 1 * 2 * 3 * 4 * 5
factorial(5) = 5 * factorial(4) = 5 * 4 * factorial(3) = 5 * 4 * 3 * factorial(2) = 5 * 4 * 3 * 2 * 1 * 1
1 \/
1 1*1
2 1 * 2
3 2 * 3
4 6 * 4
5 /|\ 24*5
|
*/
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