Q:

How to get all ranks in MPI to do a send a value to rank 0 which then does a blocking receive on all of them?

Say I have n processes:
They do a calculation, and then send the result to rank 0. This is what I want to happen:
Rank 0 will wait till it has a result from all of the ranks and then add them up.
How do I do this? Also, I want to avoid the following:
For eg. 4 processes P0, P1, P2, P3,
P1 -> P0
P2 -> P0
P3 -> P0

In the meanwhile P1 has finished its calculation and so P1->P0 happens again.
I want P0 to only make the addition for 3 processes in one cycle before doing it for the next one.
Can someone suggest an MPI function to do this? I know about MPI_Gather but I'm not sure if its blocking
I've thought up of this:
#include <mpi.h>
#include <stdio.h>

int main()
{
int pross, rank,p_count = 0;
int count = 10;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&pross);
MPI_Comm_rank(MPI_COMM_WORLD,&rank);

int * num = malloc((pross-1)*sizeof(int));

        if(rank !=0)
        {
            MPI_Send(&count,1,MPI_INT,0,1,MPI_COMM_WORLD);
        }
        else
        {            
            MPI_Gather(&count, 1,MPI_INT,num, 1, MPI_INT, 0,MPI_COMM_WORLD);
            for(ii = 0; ii < pross-1;ii++ ){printf("\n NUM %d \n",num[ii]); p_count += num[ii]; }
}
MPI_Finalize();
}

I'm getting error:
  *** Process received signal ***
  Signal: Segmentation fault (11)
  Signal code: Address not mapped (1)
  Failing at address: (nil)
  [ 0] /lib/x86_64-linux-gnu/libpthread.so.0(+0x11630)[0x7fb3e3bc3630]
  [ 1] /lib/x86_64-linux-gnu/libc.so.6(+0x90925)[0x7fb3e387b925]
  [ 2] /usr/lib/libopen-pal.so.13(+0x30177)[0x7fb3e3302177]
  [ 3] /usr/lib/libmpi.so.12(ompi_datatype_sndrcv+0x54c)[0x7fb3e3e1e3ec]
  [ 4] /usr/lib/openmpi/lib/openmpi/mca_coll_tuned.so(ompi_coll_tuned_gather_intra_basic_linear+0x143)[0x7fb3d53d9063]
  [ 5] /usr/lib/libmpi.so.12(PMPI_Gather+0x1ba)[0x7fb3e3e29a3a]
  [ 6]  sosuks(+0xe83)[0x55ee72119e83]
  [ 7] /lib/x86_64-linux-gnu/libc.so.6(__libc_start_main+0xf1)[0x7fb3e380b3f1]
  [ 8]  sosuks(+0xb5a)[0x55ee72119b5a]
  *** End of error message ***

Also, I tried:
#include <mpi.h>
#include <stdio.h>

int main()
{
int pross, rank,p_count = 0;
int count = 10;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&pross);
MPI_Comm_rank(MPI_COMM_WORLD,&rank);

int * num = malloc((pross-1)*sizeof(int));

        if(rank !=0)
        {
            MPI_Send(&count,1,MPI_INT,0,1,MPI_COMM_WORLD);
        }
        else
        {            
            MPI_Gather(&count, 1,MPI_INT,num, 1, MPI_INT, 0,MPI_COMM_WORLD);
            for(ii = 0; ii < pross-1;ii++ ){printf("\n NUM %d \n",num[ii]); p_count += num[ii]; }
}
MPI_Finalize();
}

I'm getting error here:
  *** Process received signal ***
  Signal: Segmentation fault (11)
  Signal code: Address not mapped (1)
  Failing at address: 0x560600000002
  [ 0] /lib/x86_64-linux-gnu/libpthread.so.0(+0x11630)[0x7fefc8c11630]
  [ 1] mdscisuks(+0xeac)[0x5606c1263eac]
  [ 2] /lib/x86_64-linux-gnu/libc.so.6(__libc_start_main+0xf1)[0x7fefc88593f1]
  [ 3] mdscisuks(+0xb4a)[0x5606c1263b4a]
  *** End of error message ***

For the second attempt, the thing to note here is that the send and recv were successful but root could only receive 2 messages from ranks for some reason. The segmentation fault seen was due to there being only two elements in num, I dont see why num only receives twice.
I'm calling the code as
mpiexec -n 6 ./sosuks

Can someone tell me a better / correct way to implement my idea ? 
UPDATE:
Apart from the answer below I found the mistake in my implementation above which I wanted to share:
#include <mpi.h>
#include <stdio.h>

int main()
{
int pross, rank,p_count = 0;
int count = 10;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&pross);
MPI_Comm_rank(MPI_COMM_WORLD,&rank);
MPI_Status status;
int * num = malloc((pross-1)*sizeof(int));

        if(rank !=0)
        {
            MPI_Send(&count,1,MPI_INT,0,1,MPI_COMM_WORLD);
        }
        else
        {
        int var,lick = 0;
        for(lick = 1; lick < pross; u++)
        {
        int fetihs;
        MPI_Recv(&fetihs,1,MPI_INT,lick,1,MPI_COMM_WORLD,&status);          

        var += fetihs;
        }
     // do things with var
}
MPI_Finalize();
}

A:

In your case, as Sneftel pointed out, you need MPI_Reduce. Also, you don't need explicit synchronization before the cycle completes.
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char* argv[])
{
    int pross, rank, p_count, count = 10;

    MPI_Init(&argc,&argv);
    MPI_Comm_size(MPI_COMM_WORLD, &pross);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);

    int* num = malloc((pross-1)*sizeof(int));

    // master does not send data to itself.
    // only workers send data to master.

    for (int i=0; i<3; ++i)
    {
        // to prove that no further sync is needed.
        // you will get the same answer in each cycle.
        p_count = 0; 

        if (rank == 0)
        {
            // this has not effect since master uses p_count for both 
            // send and receive buffers due to MPI_IN_PLACE.
            count = 500;

            MPI_Reduce(MPI_IN_PLACE, &p_count, 1, MPI_INT, MPI_SUM, 0, MPI_COMM_WORLD);
        }
        else
        {
            // for slave p_count is irrelevant.
            MPI_Reduce(&count, NULL, 1, MPI_INT, MPI_SUM, 0, MPI_COMM_WORLD);
        }

        if (rank == 0)
        {
            printf("p_count = %i\n", p_count);
        }

        // slaves send their data to master before the cycle completes.
        // no need for explicit sync such as MPI_Barrier.
        // MPI_Barrier(MPI_COMM_WORLD); // no need.
    }

    MPI_Finalize();
}

In the code above count in slaves are reduced to p_count in master. Note the MPI_IN_PLACE and two MPI_Reduce calls. You can get the same functionality by simply setting count = 0 and calling MPI_Reduce by all ranks without MPI_IN_PLACE.
for (int i=0; i<3; ++i)
{
    p_count = 0;    
    if (rank == 0) count = 0;

    MPI_Reduce(&count, &p_count, 1, MPI_INT, MPI_SUM, 0, MPI_COMM_WORLD);           
}

