pls convert this c++ to java code

Please help me to convert C++ program in Java code...
I am putting C++ code below

Code:

---

#include <stdio.h> // for input/output operations

#ifndef hr_timer
#include "hr_time.h"
#define hr_timer
#endif

char map[50][88]; // the map... the input file will be stored here
int island_area[20]; // array to store the various island's area
int island; // variable to store the number of islands found
int area; // temporary variable to store the + count for each island


// find a piece of land to start from.
int find(int* row, int* col)
{
int i, j;
for(i = 2; i < 83; i++)
{
for(j = 0; j < 50; j++)
{
if(map[j][i] == '+')
{
*row = j;
*col = i;
return 1;
}
}
}
return 0;
}


// look for more land, given a piece
int count(int row, int col)
{
// if it's either out of bounds or on sea, it is not land
if(row > 49 || col < 3 || col > 82 || map[row][col] != '+')
{
return 0;
}
// else, we found another bit, and will keep looking
else
{
map[row][col] = '-';
area++;

if(count(row + 1, col )) return 1; // S
if(count(row + 1, col + 1)) return 1; // SE
if(count(row , col + 1)) return 1; // E
if(count(row - 1, col + 1)) return 1; // NE
if(count(row - 1, col )) return 1; // N
if(count(row - 1, col - 1)) return 1; // NW
if(count(row , col - 1)) return 1; // W
if(count(row + 1, col - 1)) return 1; // SW
}

// if there is none around, we looked everywhere....
return 0;
}


int main()
{
int index;
FILE * input;
int row, col;

stopWatch tick;


// open the input (text)file for reading operations
input = fopen("E:\\PC2\\map.txt", "rt");
startTimer(&tick);
/* lets assume there's no problem

if(!input)
{
printf("error opening input file");
getint();
return (1);
}*/

// read the lines from the input file to a
// bidimensional array (interpreted as a map)
for(index = 0; index < 50; index++)
{
fgets(map[index], 88, input);
}

// i'm pretty sure this should be the only part being clocked,
// as this is actually where i start solving the problem itself

while(find(&row, &col)) // while there's a piece of land not scanned
{
area = 0;
count(row, col); // i'll start counting the + in it
island_area[++island] = area; // and store the count
}

printf("There are %d Continents\n", island);
for(index = 1; index < island + 1; index++)
{
printf("Continent %d has %d .\n", index, island_area[index]);
}

stopTimer(&tick);
printf("took %12.10f secs\n", getElapsedTime(&tick));

// close the input file, as it is no longer needed
fclose(input);

getchar();
return (0);
}

---

Please anyone convert this C++ program into Java as soon as possible.

Thanks in advance

Did you tried the coding in Java and are you aware about JAVA programming? Where have been stuck or you got any error message? The program that you are asking for requires a lot of practice because it is related to computer graphics. In JAVA you can also code a similar kind of program but the thing is that I would recommend you to try it on your own. We can't code it for you everything. If, however, you need help you can ask us at any point of time and we are here to help you. You can even read some books if you don't have any expertise in this fields. Some of the books which deals in Computer Graphics include "Introduction to Computer Graphics", "Computer Graphics for Java Programmers".

Our school is using JAVA, that code is our project and now , I searched in the internet about it . The problem is that the source code is c++ and I do not know how to convert it to JAVA . They never discussed the c++ in our school that's why I do not know what the code is equivalent to JAVA code .

Please make a way to convert it to java because the DEADLINE is coming .
Thank You.

By the way , that is not c++ its a C .
Sorry for the inconvenience .

#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<math.h>
int main()
{

// INPUT parameters

float power,gr,z1,z2,n1,y1,y2,ag,sigmaci,sigmas,sigmas1,sigmac1;
int zi,zs,zc;
float fw,fs,vm,Vm,ft,fd,cv,fd1,ef,fe,fo,ab,rg;
//for material properties

int ci[3][2]={{460,5000},{550,6000},{750,7500}},st[3][2]={{1350,5000},{3000,9500},{3800,11000}};

cout<<"enter the value of Z1 = ";
cin>>z1;
cout<<"enter the value of Z2 = ";
cin>>z2;
cout<<"enter the power = ";
cin>>power;
cout<<"enter the RPM of pinion = ";
cin>>n1;
gr=z2/z1;
//cout<<"the gear ratio will be = "<<gr;


cout<<"\nenter the type \n1. involute \n 2. stub teeth\n";
cin>>zi;
if(zi==1)
{
cout<<"enter the value of Angle = ";
cin>>ag;

if(ag==14.5)
{
y1=(0.124-(0.864/z1));
y2=(0.124-(0.864/z2));
fe=7850*22.5;
fo=1;
}
else if(ag==20)
{
y1=(0.154-(0.912/z1));
y2=(0.154-(0.912/z2));
fe=8150*22.5;
fo=1;
}
}
else if(zi==2)
{
y1=(0.175-(0.95/z1));
y2=(0.175-(0.95/z2));
fe=8450*22.5;
fo=0.8;
}

//for cast iron

cout<<"\nenter the grade for gear :- ";
cout<<"\n1. Grade 20 \n2. Grade 25\n3. Grade 35\n";
cin>>zc;
if(zc==1)
{
sigmaci=ci[zc][0];
sigmac1=ci[zc][1];
}
else if(zc==2)
{
sigmaci=ci[zc][0];
sigmac1=ci[zc][1];
}
else if(zc==3)
{
sigmaci=ci[zc][0];
sigmac1=ci[zc][1];
}
float P,Q;
P=sigmaci*y2;

//for steel

cout<<"\nenter the grade for pinion :- ";
cout<<"\n1. C 45 \n2. 15 Ni 2 Cr 1 Mo 15\n3. 40 Ni 2 Cr 1 Mo 28\n";
cin>>zs;
if(zs==1)
{
sigmas=st[zs][0];
sigmas1=st[zs][1];

}
else if(zs==2)
{
sigmas=st[zs][0];
sigmas1=st[zs][1];
}
else if(zs==3)
{
sigmas=st[zs][0];
sigmas1=st[zs][1];
}
Q=sigmas*y1;

// Design basis


double a,b=0,c,d,e,f,g,h,i,j,k,l,m,n,p,r,s,t,u,x1,x2,x3;

int w;
printf("\nThe cubic equation is of type a*x^3+b*x^2+c*x+d \n");
if(P<Q)
{
cout<<"\n Design for strength and check for wear \n";

a=39.47*P;
c=(-25)*power;
d=((-23.87)*power)/(n1*z1);
e=2.7182818284590;
f=((3*c/a)-(b*b/(a*a)))/3;
g=((2*b*b*b/(a*a*a))-(9*b*c/(a*a))+(27*d/a))/27;
h=(g*g/4)+(f*f*f/27);
i=sqrt(((g*g/4)-h));
j=exp(log10(i)/log10(e)/3);
k=acos((-1)*(g/(2*i)));
l=j*(-1);
m=cos(k/3);
n=sqrt(3)*sin(k/3);
p=(b/3*a)*(-1);
r=(-1)*(g/2)+sqrt(h);
s=exp(log10(r)/log10(e)/3);
t=(-1)*(g/2)-sqrt(h);
u=exp(log10(t)/log10(e)/3);
if (h>0) w=1;
if (h<=0) w=3;
if ((f==0) && (g==0) && (h==0)) w=2;
switch (w)
{
case 1:
x1=(s+u)-(b/3*a);
x2=(-1)*(s+u)/2-(b/3*a);
x3=(s-u)*sqrt(3)/2;
//cout<<"x1 = "<<x1;
printf("\nA 3 pont:\n%lf\n%lf +i*%lf\n%lf -i*%lf", x1, x2, x3, x2, x3);
break;
case 2:
x1=exp(log10(d/a)/log10(e)/3)*(-1);
printf("\n\n%lf", x1);
break;
case 3:
x1=2*j*cos(k/3)-(b/3*a);
x2=l*(m+n)+p;
x3=l*(m-n)+p;
//cout<<"x1 = "<<x1;
printf("\nMODULE are :- \n%lf\n%lf\n%lf", x1, x2, x3);
break;
}



}
else
{
cout<<"\n Design for wear and check for strength \n";

a=(1.2)*(z1*z1)*n1*gr*sigmas1*sigmas1;
c=(1414*1000)*(gr+1)*n1*z1*(-1)*power;
d=(-1)*8.1*power*(gr+1)*pow(10,9);

e=2.7182818284590;
f=((3*c/a)-(b*b/(a*a)))/3;
g=((2*b*b*b/(a*a*a))-(9*b*c/(a*a))+(27*d/a))/27;
h=(g*g/4)+(f*f*f/27);
i=sqrt(((g*g/4)-h));
j=exp(log10(i)/log10(e)/3);
k=acos((-1)*(g/(2*i)));
l=j*(-1);
m=cos(k/3);
n=sqrt(3)*sin(k/3);
p=(b/3*a)*(-1);
r=(-1)*(g/2)+sqrt(h);
s=exp(log10(r)/log10(e)/3);
t=(-1)*(g/2)-sqrt(h);
u=exp(log10(t)/log10(e)/3);
if (h>0) w=1;
if (h<=0) w=3;
if ((f==0) && (g==0) && (h==0)) w=2;
switch (w)
{
case 1:
x1=(s+u)-(b/3*a);
x2=(-1)*(s+u)/2-(b/3*a);
x3=(s-u)*sqrt(3)/2;
//cout<<"x1 = "<<x1;
printf("\nMODULE are :- \n%lf\n%lf +i*%lf\n%lf -i*%lf", x1, x2, x3, x2, x3);
break;
case 2:
x1=exp(log10(d/a)/log10(e)/3)*(-1);
printf("\n\n%lf", x1);
break;
case 3:
x1=2*j*cos(k/3)-(b/3*a);
x2=l*(m+n)+p;
x3=l*(m-n)+p;
//cout<<"x1 = "<<x1;
printf("\nA 3 pont:\n%lf\n%lf\n%lf", x1, x2, x3);
break;
}


}

if(x1<0.1||x1>2)
{
cout<<"\n ERROR !!!! ";
}
else
{
int mec=1;
while(mec==1)
{

x1=x1*10;
float table[14]={1,1.25,1.5,2,2.5,3,4,5,6,8,10,12,16,20},nm;
for(int y=1;y<=13;y++)
{
if(x1>table[y-1]&&x1<=table[y])
{
x1=table[y];
nm=table[y+1];

break;
}
}

//cout<<" new x1 = "<<x1;


x1=x1/10;
vm=(3.14*n1*x1*z1)/6000;
Vm=vm*60;
ef=fe/Vm;
ft=(power*75)/vm;
cv=(1+(vm/3));
fd=ft*cv;
ab=0.164*Vm*((ef*4*3.14*x1)+ft);
rg=(0.164*Vm)+(1.485*(sqrt((ef*4*3.14*x1)+ft)));
fd1=ft+(ab/rg);
printf("\n BUCKINGHAM DYNAMIC LOAD = %lfkgf",fd1);

cout<<"\n new module = "<<x1;
if(P<Q)
{
fw=(sigmas1*sigmas1*1.685*sin(ag)*x1*z1*gr)/(1000000*(gr+1));
cout<<"\n wear force Fw = "<<fw<<" KgF";
}
else
{
fs=4*Q*3.14159*3.14159*x1*x1;
cout<<"\n Fs = "<<fs<<" KgF";
}

if(P<Q)
{

if(fw>=fd&&fw>=fd1)
{
cout<<"\n SAFE ";
mec=2;
}
else
{
cout<<"\n UNSAFE ";
mec=1;
nm=nm/10;
x1=nm;
}
}
else
{
if(fs>=fd&&fs>=fd1)
{
cout<<"\n SAFE ";
mec=2;
}
else
{
cout<<"\n UNSAFE ";
mec=1;
nm=nm/10;
x1=nm;
}
}
}

printf("\n BASIC DIMENSIONS:-\n");
float module;
module=x1*10;
printf("\n Module= %lfmm",module);
float cd;
cd=(0.5*module*(z1+z2));
printf("\n Centre Distance = %lfmm",cd);
float bc,pd1,pd2,td,da1,da2,df1,df2;
if(fo==1)
{
bc=0.25*module;
}
else
{
bc=0.3*module;
}
printf("\n Bottom Clearance = %lfmm",bc);
if(fo==1)
{
td=2.25*module;
}
else
{
td=1.9*module;
}
printf("\n Tooth Depth = %lfmm",td);
pd1=module*z1;
pd2=module*z2;
printf("\n Pitch Diameter di= %lfmm",pd1);
printf("\n Pitch Diameter d2= %lfmm",pd2);
da1=(z1+(2*fo))*module;
da2=(z2+(2*fo))*module;
df1=(((z1-(2*fo))*module)-(2*bc));
df2=(((z2-(2*fo))*module)-(2*bc));
printf("\n Tip Diameter da1 = %lfmm",da1);
printf("\n Tip Diameter da2 = %lfmm",da2);
printf("\n Root Diameter df1 = %lfmm",df1);
printf("\n Root Diameter df2 = %lfmm",df2);
}
getch();
return(0);
}