Reducing the max value and saving its index

TL;DR You can use User-Defined Reduction.

First, instead of:

for(i = 0; i< 10; i++){
    if (v[i] > maximo)
        maximo = v[i];
    b = i + 100;
}

you meant this:

for(i = 0; i< 10; i++){
    if (v[i] > maximo){
        maximo = v[i];
        b = i + 100;
    }
}

OpenMP has in-build reduction functions that consider a single target value, however in your case you want to reduce taking into account 2 values the max and the array index. After OpenMP 4.0 one can create its own reduction functions (i.e., User-Defined Reduction).

First, create a struct to store the two relevant values:

struct MyMax {
   int max;
   int index;
};

then we need to teach the OpenMP implementation how to reduce it:

#pragma omp declare reduction(maximo : struct MyMax : omp_out = omp_in.max > omp_out.max ? omp_in : omp_out)

we set our parallel region accordingly:

    #pragma omp parallel for reduction(maximo:myMaxStruct)
    for(int i = 0; i< 10; i++){
       if (v[i] > myMaxStruct.max){
          myMaxStruct.max = v[i];
          myMaxStruct.index = i + 100;
      }
   }

Side Note You do not really need private(i), because with the #pragma omp parallel for the index variable of the for loop will be implicitly private anyway.

All put together:

#include <stdio.h>
#include <stdlib.h>
#include <omp.h>

struct MyMax {
   int max;
   int index;
};


int main(void)
{
    #pragma omp declare reduction(maximo : struct MyMax : omp_out = omp_in.max > omp_out.max ? omp_in : omp_out)
    struct MyMax myMaxStruct;
    myMaxStruct.max = 0;
    myMaxStruct.index = 0;

    int v[10] = {2,9,1,3,5,7,1,2,0,0};

    #pragma omp parallel for reduction(maximo:myMaxStruct)
    for(int i = 0; i< 10; i++){
       if (v[i] > myMaxStruct.max){
          myMaxStruct.max = v[i];
          myMaxStruct.index = i + 100;
      }
   }
   printf("Max %d : Index %d\n", myMaxStruct.max, myMaxStruct.index);
}

OUTPUT:

Max 9 : Index 101

(Index is 101 because you have b = i + 100)

I've coded this but not compiled or tested it:

int v[10] = { 2, 9, 1, 3, 5, 7, 1, 2, 0, 0 };

int maximo = 0;
int b = 0;
int i;

int nt = omp_get_num_threads();
int bv[nt] = { 0 };

#pragma omp parallel for shared(v) shared(bv) private(i) reduction(max:maximo)
for (i = 0; i < 10; i++) {
    if (v[i] > maximo) {
        maximo = v[i];
        bv[omp_get_thread_num()] = i + 100;
    }
}

for (i = 0;  i < nt;  ++i)
    printf("bv[%d] = %d\n",i,bv[i]);

Beware that "Returns the number of threads in the current team. In a sequential section of the program omp_get_num_threads returns 1"

Okay, I've recoded it [and built/run it] and it does produce one non-zero bv output:

#include <stdio.h>
#include <omp.h>

int
main(void)
{
    int v[10] = { 2, 9, 1, 3, 5, 7, 1, 2, 0, 0 };

    int i;
    int nt;
    int maximo = 0;
    int index = 0;
    int bv[32] = { 0 };
    int max[32] = { 0 };
    #pragma omp parallel shared(v, bv)
    {
        nt = omp_get_num_threads();
        int thread_id = omp_get_thread_num();
        #pragma omp for private(i)
        for (i = 0; i < 10; i++) {
            if (v[i] > max[thread_id]) {
               max[thread_id] = v[i];
               bv[thread_id] = i + 100;
            }
        }
    }
    // Reducing sequentially 
    for (i = 0;  i < nt;  ++i){
        if(max[i] > maximo){
           maximo = max[i];
           index  = bv[i];
        }
    }
    printf("Max %d at index %d\n", maximo, index);
    return 0;
}

Here is the program output:

Max 9 at index 101