simulation algorithm implementation in C/C++ - c

Question

Let a row of 8000 lamps. Initially, only the one located to the left is lit.
Then, every second, the following operation is performed: each lamp changes state (on or off) if the one on its left was lit a second before. The leftmost lamp stays on all the time. This operation is instantaneous.
The process stops when the lamp at the right end lights for the first time.
How many lights are on?
My following implementation of the problem is false, can you help me?
#include <cstdio>
int t[8001][2];
int main()
{
t[1][0] = 1;
t[1][1] = 1;
int cpt1 = 0, ip = 0;
while (t[8000][0] != 1 && t[8000][1] != 1)
{
ip++;
for (int j=2;j<8001;j++)
{
if(t[j-1][!(ip&1)])
t[j][(ip & 1)] = !t[j][!(ip & 1)];
}
}
for(int j = 1;j < 8001; j++)
cpt1 += t[j][1];
printf("cpt=%d\n", cpt1);
}

Answer 1

Code is missing an update when the left does not change.
Code simplified (zero based offset, use of bool) and corrected below
#include<stdbool.h>
#include<stdio.h>
#define N 8000
bool t[N][2];
int main(void) {
t[0][0] = true;
t[0][1] = true;
int ip = 0;
while (t[N - 1][0] == 0 && t[N - 1][1] == 0) {
ip = !ip;
for (int j = 1; j < N; j++) {
if (t[j - 1][!ip]) {
t[j][ip] = !t[j][!ip];
} else {
t[j][ip] = t[j][!ip]; // add
}
}
}
int cpt1 = 0;
for (int j = 0; j < N; j++) {
cpt1 += t[j][1];
}
printf("N=%d cpt=%d\n", N, cpt1);
return 0;
}
Output
N=8000 cpt=2048

Answer 2

the following proposed code:
cleanly compiles
uses C header files rather than C++ header files
performs the desired operation, but not the fastest possible algorithm
is liberally commented
And now the proposed code:
#include <stdio.h>
int t1[8000]; // initially all zeros
int t2[8000];
int main( void )
{
// setup initial conditions
int numLitLights = 0;
t1[0] = 1;
// while stop condition not true
while ( t1[7999] != 1 )
{
// make one pass through lamps
// update values
for (int j=0; j<7999; j++)
{
if( t1[j] )
{
t2[j+1] = ( t1[j+1] )? 0 : 1;
}
}
// update original
for( int j=0; j< 8000; j++ )
{
t1[j] = t2[j];
}
}
// count lit lamps
for(int j = 0; j < 8000; j++)
{
if( t1[j] )
{
numLitLights++;
}
}
// output number of lit lamps
printf( "number of lit lamps: %d\n", numLitLights );
} // end function: main
The result (number of lamps lit) is
1024