got your idea thanks.
I was also thinking this way but couldn’t come with a%=3,b%=3;
this step.
Here is my solution.
if(a<b) swap(a,b);
if(a>2*b)
{
cout<<"NO" << endl;
return 0;
}
if((a+b)%3==0)
{
cout<<"YES"<< endl;
}
else cout<<"NO"<< endl;
I know this is from a while ago, but i thought there wasn’t sufficient explanations in any of the other answers.
Lets say x times we take 2 from a , and 1 from b
and y times we take 2 from b and 1 from a
meaning:
a = 2x + 1y
b = 1x + 2y
on solving for a and b
2a - b = 3x
2b - a = 3y
x and y should be non negative , and from here we can derive (a+b)%3==0.
how did you come up?
since he swaps a and b if a is less than b,so a must be atmost 2b if we want to empty both piles as 2 coins from a and 1 coin from b will be used but if a > 2b then whole pile of b will become empty and some coins will still be left in a.
Let’s assume that we are removing x number of times 2 coins from pile a and 1 coin from pile b.
Let’s assume that we are removing y number of times 1 coin from pile a and 2 coins from pile b.
Mathematically we can write our assumptions as:
a = 2x + y \;\;\cdots (i)
b = x + 2y \;\; \cdots (ii)
We have two equations and we have two unknowns, means we can solve for x and y.
After solving the above equations for x and y, you will get:
x = \frac{2a - b}{3}
y = \frac{2b - a}{3}
Clearing x and y \in Z^{+} and they must be a multiple of 3.
One more important information we will get when we subtract (ii) from (i) i.e.
|a - b| = |x - y|
Source Code: Competitive-Programming/CSES/Coin-Piles at master · strikersps/Competitive-Programming · GitHub
My solution.
#include<iostream>
void swap(int *,int *);
int main(){
int t, a, b, x;
std::cin >> t;
while ( t-- ) {
bool flag = false;
std::cin >> a >> b;
if (b>a) swap(&a , &b) ;
x = a-b;
if (x > b) flag = true;
a -= 2 * x;
b -= x;
if (a != b) flag = true;
if (a % 3 != 0) flag = true;
if (flag) std::cout << "NO\n" ;
else std :: cout << "YES\n" ;
}
return 0;
}
void swap (int *a, int *b){
int temp = *a;
*a = *b;
*b = temp;
}
I am confused by what you mean when you say the following
I do not understand what you mean by x number of times and y number of times, and also I am confused as to why taking 1 coin from pile b affects the a equation (i.e. a = 2x + y).
Any sort of clarification would be very helpful. Thank you.
This is what I did :
void solve(){
ll a, b;
cin >> a >> b;
if ((a + b) % 3 == 0 && 2 * a >= b && 2 * b >= a )
{
cout << "YES";
}
else cout << "NO";
}
It means …When we reduce “a” by 2 then we must reduce “b” by 1 and when we reduce “a” by 1 the we must reduce “b” by 2. So if We assume that we take x times 2 and y times 1 to reach a to 0 then we must take x times 1 and y times 2 to reduce “b” to 0. i.e
a=2x+1y
b=1x+2y
Thanks for the clear explanation!!
My sol:
#include<bits/stdc++.h>
typedef long long int ll;
using namespace std;
int main()
{
long long unsigned int t,a,b;
cin>>t;
while(t--)
{
cin>>a>>b;
if((2*a-b)%3==0 and (2*b-a)%3==0)
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
}
return 0;
}
Thank you for explaining it in detail
thanks">
?>
why you are doing this
if(a>2*b){
cout<<"NO\n";
}
Hey, In the first line, he swaps a,b if a is smaller. So now, a will always be greater than equal to b.
In the worst case, if a is very large (as compared to b). We will always remove 2 from a and 1 from b (as a is the larger one).
eg - 24 12.
but if a is greater than 24 in this case - say 25,26,100 etc and b is 12.
The best you can do is keep removing 2 from a and 1 from b. but if a > 24, b will become 0 and a will still be nonzero.
if((a+b)%3==0)
how is this true
I solved it using binary search, but it seems like it had a way simpler solution.
void solve() {
ll n, temp, a, b, ans = 0;
cin >> a >> b;
if(a == 2 * b || b == 2 * a) {
cout << "YES" << endl;
return;
}
if(a == 0 || b == 0) {
cout << "NO" << endl;
return;
}
ll start = 0, end = max(a, b);
while(start < end) {
ll mid = (start + end)/2;
if(a - mid == 2 * (b - 2 * mid) && a - mid > 0) {
cout << "YES" << endl;
return;
}
if(a - mid > 2 * (b - 2 * mid)) end = mid;
else start = mid + 1;
}
cout << "NO" << endl;
}
@prashant2020
let’s think about extreme condition , everytime you take 2 coin from a and 1 from b .
then also, a ratio b will be exact 2;
or else, their ratio will be less than 2; hence, it can never succeed 2;
Hope it helps.