Computer Guesses Your Number
I need to make a "Guessing game" but I'm stuck because I'm not sure if this is the right way to make it, since it's so hard to actually find the number I'm thinking of when playing the game.
#include <iostream>
#include <string>
using namespace std;
int main ()
{
int guess;
int n=500;
cout<<"Think of a number from 1 to 1000."<<endl;
cout<<"The number is : 1. 500"<<endl;
cout<<"2. Bigger than 500"<<endl;
cout<<"3. Smaller than 500"<<endl;
cin>>guess;
while(n<=2000)
{
if(guess==1)
{
cout<<"The computer has guessed the number!";
break;
}
else if(guess==2)
{
n+=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
else if(guess==3)
{
n-=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
}
}
I can't seem to find a better way to make the game. The number always goes over 1000
so I had to set the while
to under 2000
for it to work.
It would be good if I could set the n+=n/2
formula so that it splits the last number that was the n
variable. For example : 500+(500/2)
, then 750+(250/2)
, then 875+(125/2)
(not sure how it would continue when the 125
will get split into 62.5
) or the same with the n-=n/2
.
c++
add a comment |
I need to make a "Guessing game" but I'm stuck because I'm not sure if this is the right way to make it, since it's so hard to actually find the number I'm thinking of when playing the game.
#include <iostream>
#include <string>
using namespace std;
int main ()
{
int guess;
int n=500;
cout<<"Think of a number from 1 to 1000."<<endl;
cout<<"The number is : 1. 500"<<endl;
cout<<"2. Bigger than 500"<<endl;
cout<<"3. Smaller than 500"<<endl;
cin>>guess;
while(n<=2000)
{
if(guess==1)
{
cout<<"The computer has guessed the number!";
break;
}
else if(guess==2)
{
n+=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
else if(guess==3)
{
n-=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
}
}
I can't seem to find a better way to make the game. The number always goes over 1000
so I had to set the while
to under 2000
for it to work.
It would be good if I could set the n+=n/2
formula so that it splits the last number that was the n
variable. For example : 500+(500/2)
, then 750+(250/2)
, then 875+(125/2)
(not sure how it would continue when the 125
will get split into 62.5
) or the same with the n-=n/2
.
c++
4
if the number is known to be over 500, you never want to select a 500 or less ever again. Store a min and a max and keep picking numbers between the min and max, moving the min or max when the user provides new information.
– user4581301
Nov 23 '18 at 22:48
@edlars47. Read this example: guessing game for inspiration
– eapetcho
Nov 23 '18 at 22:57
2
This isn't a C++ problem, it's a logic problem. I recommend pen, paper and patience!
– Lightness Races in Orbit
Nov 23 '18 at 23:04
Look up "binary search". You're on the right track but your implementation is flawed.
– Fei Xiang
Nov 23 '18 at 23:29
add a comment |
I need to make a "Guessing game" but I'm stuck because I'm not sure if this is the right way to make it, since it's so hard to actually find the number I'm thinking of when playing the game.
#include <iostream>
#include <string>
using namespace std;
int main ()
{
int guess;
int n=500;
cout<<"Think of a number from 1 to 1000."<<endl;
cout<<"The number is : 1. 500"<<endl;
cout<<"2. Bigger than 500"<<endl;
cout<<"3. Smaller than 500"<<endl;
cin>>guess;
while(n<=2000)
{
if(guess==1)
{
cout<<"The computer has guessed the number!";
break;
}
else if(guess==2)
{
n+=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
else if(guess==3)
{
n-=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
}
}
I can't seem to find a better way to make the game. The number always goes over 1000
so I had to set the while
to under 2000
for it to work.
It would be good if I could set the n+=n/2
formula so that it splits the last number that was the n
variable. For example : 500+(500/2)
, then 750+(250/2)
, then 875+(125/2)
(not sure how it would continue when the 125
will get split into 62.5
) or the same with the n-=n/2
.
c++
I need to make a "Guessing game" but I'm stuck because I'm not sure if this is the right way to make it, since it's so hard to actually find the number I'm thinking of when playing the game.
#include <iostream>
#include <string>
using namespace std;
int main ()
{
int guess;
int n=500;
cout<<"Think of a number from 1 to 1000."<<endl;
cout<<"The number is : 1. 500"<<endl;
cout<<"2. Bigger than 500"<<endl;
cout<<"3. Smaller than 500"<<endl;
cin>>guess;
while(n<=2000)
{
if(guess==1)
{
cout<<"The computer has guessed the number!";
break;
}
else if(guess==2)
{
n+=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
else if(guess==3)
{
n-=n/2;
cout<<"The number is :n1."<<n<<endl;
cout<<"2. Bigger than "<<n<<endl;
cout<<"3. Smaller than "<<n<<endl;
cin>>guess;
}
}
}
I can't seem to find a better way to make the game. The number always goes over 1000
so I had to set the while
to under 2000
for it to work.
It would be good if I could set the n+=n/2
formula so that it splits the last number that was the n
variable. For example : 500+(500/2)
, then 750+(250/2)
, then 875+(125/2)
(not sure how it would continue when the 125
will get split into 62.5
) or the same with the n-=n/2
.
c++
c++
edited Nov 23 '18 at 22:50
bcperth
2,0011514
2,0011514
asked Nov 23 '18 at 22:42
edlars47edlars47
32
32
4
if the number is known to be over 500, you never want to select a 500 or less ever again. Store a min and a max and keep picking numbers between the min and max, moving the min or max when the user provides new information.
– user4581301
Nov 23 '18 at 22:48
@edlars47. Read this example: guessing game for inspiration
– eapetcho
Nov 23 '18 at 22:57
2
This isn't a C++ problem, it's a logic problem. I recommend pen, paper and patience!
– Lightness Races in Orbit
Nov 23 '18 at 23:04
Look up "binary search". You're on the right track but your implementation is flawed.
– Fei Xiang
Nov 23 '18 at 23:29
add a comment |
4
if the number is known to be over 500, you never want to select a 500 or less ever again. Store a min and a max and keep picking numbers between the min and max, moving the min or max when the user provides new information.
– user4581301
Nov 23 '18 at 22:48
@edlars47. Read this example: guessing game for inspiration
– eapetcho
Nov 23 '18 at 22:57
2
This isn't a C++ problem, it's a logic problem. I recommend pen, paper and patience!
– Lightness Races in Orbit
Nov 23 '18 at 23:04
Look up "binary search". You're on the right track but your implementation is flawed.
– Fei Xiang
Nov 23 '18 at 23:29
4
4
if the number is known to be over 500, you never want to select a 500 or less ever again. Store a min and a max and keep picking numbers between the min and max, moving the min or max when the user provides new information.
– user4581301
Nov 23 '18 at 22:48
if the number is known to be over 500, you never want to select a 500 or less ever again. Store a min and a max and keep picking numbers between the min and max, moving the min or max when the user provides new information.
– user4581301
Nov 23 '18 at 22:48
@edlars47. Read this example: guessing game for inspiration
– eapetcho
Nov 23 '18 at 22:57
@edlars47. Read this example: guessing game for inspiration
– eapetcho
Nov 23 '18 at 22:57
2
2
This isn't a C++ problem, it's a logic problem. I recommend pen, paper and patience!
– Lightness Races in Orbit
Nov 23 '18 at 23:04
This isn't a C++ problem, it's a logic problem. I recommend pen, paper and patience!
– Lightness Races in Orbit
Nov 23 '18 at 23:04
Look up "binary search". You're on the right track but your implementation is flawed.
– Fei Xiang
Nov 23 '18 at 23:29
Look up "binary search". You're on the right track but your implementation is flawed.
– Fei Xiang
Nov 23 '18 at 23:29
add a comment |
1 Answer
1
active
oldest
votes
#include <iostream>
using namespace std;
int main()
{
int guess;
int min = 0, current = 500, max = 1000;
cout << "Think of a number from 1 to 1000." << endl;
cout << "The number is : 1. 500" << endl;
cout << "2. Bigger than 500" << endl;
cout << "3. Smaller than 500" << endl;
cin >> guess;
while (current <= 2000)
{
if (guess == 1)
{
cout << "The computer has guessed the number!";
break;
}
else if (guess == 2)
{
min = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
else if (guess == 3)
{
max = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
}
}
This should work. the min
and max
effectively restrict the domain after each guessing.
also, int variable doesn't have decimals. it will round if an odd number is divided by 2.
– Richard w
Nov 23 '18 at 23:32
while (current <= 2000)
is useless as an exit condition.
– user4581301
Nov 24 '18 at 0:48
yep, forgot to delete that
– Richard w
Nov 24 '18 at 3:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
#include <iostream>
using namespace std;
int main()
{
int guess;
int min = 0, current = 500, max = 1000;
cout << "Think of a number from 1 to 1000." << endl;
cout << "The number is : 1. 500" << endl;
cout << "2. Bigger than 500" << endl;
cout << "3. Smaller than 500" << endl;
cin >> guess;
while (current <= 2000)
{
if (guess == 1)
{
cout << "The computer has guessed the number!";
break;
}
else if (guess == 2)
{
min = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
else if (guess == 3)
{
max = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
}
}
This should work. the min
and max
effectively restrict the domain after each guessing.
also, int variable doesn't have decimals. it will round if an odd number is divided by 2.
– Richard w
Nov 23 '18 at 23:32
while (current <= 2000)
is useless as an exit condition.
– user4581301
Nov 24 '18 at 0:48
yep, forgot to delete that
– Richard w
Nov 24 '18 at 3:01
add a comment |
#include <iostream>
using namespace std;
int main()
{
int guess;
int min = 0, current = 500, max = 1000;
cout << "Think of a number from 1 to 1000." << endl;
cout << "The number is : 1. 500" << endl;
cout << "2. Bigger than 500" << endl;
cout << "3. Smaller than 500" << endl;
cin >> guess;
while (current <= 2000)
{
if (guess == 1)
{
cout << "The computer has guessed the number!";
break;
}
else if (guess == 2)
{
min = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
else if (guess == 3)
{
max = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
}
}
This should work. the min
and max
effectively restrict the domain after each guessing.
also, int variable doesn't have decimals. it will round if an odd number is divided by 2.
– Richard w
Nov 23 '18 at 23:32
while (current <= 2000)
is useless as an exit condition.
– user4581301
Nov 24 '18 at 0:48
yep, forgot to delete that
– Richard w
Nov 24 '18 at 3:01
add a comment |
#include <iostream>
using namespace std;
int main()
{
int guess;
int min = 0, current = 500, max = 1000;
cout << "Think of a number from 1 to 1000." << endl;
cout << "The number is : 1. 500" << endl;
cout << "2. Bigger than 500" << endl;
cout << "3. Smaller than 500" << endl;
cin >> guess;
while (current <= 2000)
{
if (guess == 1)
{
cout << "The computer has guessed the number!";
break;
}
else if (guess == 2)
{
min = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
else if (guess == 3)
{
max = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
}
}
This should work. the min
and max
effectively restrict the domain after each guessing.
#include <iostream>
using namespace std;
int main()
{
int guess;
int min = 0, current = 500, max = 1000;
cout << "Think of a number from 1 to 1000." << endl;
cout << "The number is : 1. 500" << endl;
cout << "2. Bigger than 500" << endl;
cout << "3. Smaller than 500" << endl;
cin >> guess;
while (current <= 2000)
{
if (guess == 1)
{
cout << "The computer has guessed the number!";
break;
}
else if (guess == 2)
{
min = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
else if (guess == 3)
{
max = current;
current = (min + max) / 2;
cout << "The number is :n." << current << endl;
cout << "2. Bigger than " << current << endl;
cout << "3. Smaller than " << current << endl;
cin >> guess;
}
}
}
This should work. the min
and max
effectively restrict the domain after each guessing.
edited Nov 24 '18 at 0:48
user4581301
19.6k51831
19.6k51831
answered Nov 23 '18 at 23:31
Richard wRichard w
16
16
also, int variable doesn't have decimals. it will round if an odd number is divided by 2.
– Richard w
Nov 23 '18 at 23:32
while (current <= 2000)
is useless as an exit condition.
– user4581301
Nov 24 '18 at 0:48
yep, forgot to delete that
– Richard w
Nov 24 '18 at 3:01
add a comment |
also, int variable doesn't have decimals. it will round if an odd number is divided by 2.
– Richard w
Nov 23 '18 at 23:32
while (current <= 2000)
is useless as an exit condition.
– user4581301
Nov 24 '18 at 0:48
yep, forgot to delete that
– Richard w
Nov 24 '18 at 3:01
also, int variable doesn't have decimals. it will round if an odd number is divided by 2.
– Richard w
Nov 23 '18 at 23:32
also, int variable doesn't have decimals. it will round if an odd number is divided by 2.
– Richard w
Nov 23 '18 at 23:32
while (current <= 2000)
is useless as an exit condition.– user4581301
Nov 24 '18 at 0:48
while (current <= 2000)
is useless as an exit condition.– user4581301
Nov 24 '18 at 0:48
yep, forgot to delete that
– Richard w
Nov 24 '18 at 3:01
yep, forgot to delete that
– Richard w
Nov 24 '18 at 3:01
add a comment |
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4
if the number is known to be over 500, you never want to select a 500 or less ever again. Store a min and a max and keep picking numbers between the min and max, moving the min or max when the user provides new information.
– user4581301
Nov 23 '18 at 22:48
@edlars47. Read this example: guessing game for inspiration
– eapetcho
Nov 23 '18 at 22:57
2
This isn't a C++ problem, it's a logic problem. I recommend pen, paper and patience!
– Lightness Races in Orbit
Nov 23 '18 at 23:04
Look up "binary search". You're on the right track but your implementation is flawed.
– Fei Xiang
Nov 23 '18 at 23:29