11489 - Integer Game

[aha2007]

Anyone give me any hints:) Really cool Problem

[andmej]

There's an O(n) solution.

Invisible text follows:

First determine if the first player can remove a number.
If he can, then notice than from now on the only possible numbers than can be removed are 3, 6, or 9. Also notice that it's completely equivalent to delete any removable number, because you won't get any extra advantage if you choose 9 instead of 3 or whatever. The only thing that matters is that it must be divided evenly by 3.
From this, you can count the number of removable digits and determine the winner by the parity of the count.


Good luck!

[peterkwan]

I am getting WA all the time. Please help to give some critical inputs. Thanks.

Code: Select all

#include <iostream>
using namespace std;

int main() {
  int n;
  cin >> n;
  for (int i=0; i<n; i++) {
    string s;
    cin >> s;
    int n2 = s.size();
    cout << "Case " << (i+1) << ": ";
 //   cout << s << " - ";

    if (n2==1)
      cout << "S" << endl;
    else if (n2==2)
      cout << "T" << endl;
    else {
      int sum = 0, n1 = 0;
      for (int j=0;j<n2;j++) {
        int c=(s[j]-'0');
        sum+=c;
        if (c%3==0) n1++;
      }
      if (n1==n2) {
        if (n1%2==0) 
          cout << "T" << endl;
        else
          cout << "S" << endl;
      }
      else if (sum%3==0) {
        if (n1%2==0) 
          cout << "T" << endl;
        else
          cout << "S" << endl;
      }
      else if (n2-n1<=2) {
        if (n1%2==0) 
          cout << "T" << endl;
        else
          cout << "S" << endl;
      }
      else if ((n2-n1)%3==2) {
        if (n1%2==0) 
          cout << "T" << endl;
        else
          cout << "S" << endl;
      }
      else {
        if (n1%2==0) 
          cout << "S" << endl;
        else
          cout << "T" << endl;
      }
    }
  }
  return 0;
}

[LayCurse]

I am getting WA all the time. Please help to give some critical inputs. Thanks.

Try this.
input:

6
443
4444
44444
44455
44555
45555

output:

Case 1: T
Case 2: S
Case 3: T
Case 4: S
Case 5: S
Case 6: T

[peterkwan]

LayCurse, thanks for your input. I modified my code and finally passed the test case above. However, I am still getting WA. Could somebody helps what test case my code below will fail? Thanks a lot.

Code: Select all

#include <iostream>
using namespace std;

main() {
  int n;
  cin >> n;
  for (int i=0; i<n; i++) {
    string s;
    cin >> s;
    int n2 = s.size();
    cout << "Case " << (i+1) << ": ";
 //   cout << s << " - ";

    if (n2==1)
      cout << "S" << endl;
    else if (n2==2)
      cout << "T" << endl;
    else {
      int sum = 0, n1 = 0;
      string s1="", s2="";
      for (int j=0;j<n2;j++) {
        int c=(s[j]-'0');
        if (c%3==0) n1++;
        else { 
          if (c%3==1)
            s1+=s[j];
          else
            s2+=s[j];
          sum+=c;
        }      
      }
      if (n1==n2) {
        if (n1%2==0) 
          cout << "T" << endl;
        else
          cout << "S" << endl;
      }
      else if (sum%3==0) {
        if (n1%2==0) 
          cout << "T" << endl;
        else
          cout << "S" << endl;
      }
      else if ((n2-n1)==1) {
        if (n1%2==0) 
          cout << "T" << endl;
        else
          cout << "S" << endl;
      }
      else if ((n2-n1)==2) {
        cout << "T" << endl;
      }
      else {
        if (sum%3==2 && s2.size()>=1) {
          if (n1%2==0)
            cout << "S" << endl;
          else
            cout << "T" << endl;
        }
        else if (sum%3==1 && s1.size()>=1) {
          if (n1%2==0)
            cout << "S" << endl;
          else
            cout << "T" << endl;
        }
        else {
          if (n1%2==0)
            cout << "T" << endl;
          else
            cout << "S" << endl;
        }
      }
    }
  }
  return 0;
}

[Obaida]

Some one can explain these case???
INPUT:

Code: Select all

2
44455
44555

OUTPUT:

Code: Select all

S
S

Why???

The problem statement says if the number of digits in N is odd then S wins otherwise T wins.

I am confused.

[peterkwan]

Some one can explain these case???
INPUT:

Code: Select all

2
44455
44555

OUTPUT:

Code: Select all

S
S

Why???
I am confused.

This is because:

A player can remove a particular digit if the sum of digits of the resulting number is a multiple of 3 or there are no digits left.

Since S moves first, the obvious way is to remove the redundant digit (4 in the first case and 5 in the second case). After removing the digit, there is no way for T to remove a digit to remain the resulting sum to be a multiple of 3. Therfore, S wins.

[sohel]

Some one can explain these case???
INPUT:

Code: Select all

2
44455
44555

OUTPUT:

Code: Select all

S
S

Why???

The problem statement says if the number of digits in N is odd then S wins otherwise T wins.

I am confused.

You should continue reading on from that sentence. The next sentence that follows goes like this : "To make the game more interesting, we apply one additional constraint. A player can remove a particular digit if the sum of digits of the resulting number is a multiple of 3 or there are no digits left."

Hope it clears up things!

[ssavi]

Why It is Getting WA?? I have Checked All The Tests of Samples , uDebug and OJ Borads . All are Ok . But I am getting WA . Here is my Code: -

Code: Select all

#include<bits/stdc++.h>

using namespace std;

int main()
{
    int t, caseno=0;
    char a[1005];
    scanf("%d",&t);
    getchar();
    for(int k=0; k<t; k++)
    {
        gets(a);
        int digsum = 0, res;
        for(int i=0; i<strlen(a); i++)
            { digsum = digsum + (a[i]-'0'); }
        int s = 0;
        int t = 0;
        for(int i=0; i<strlen(a); i++)
        {
            if(a[i]!='0')
            {
                res = digsum - (a[i]-'0');
                if(res%3==0)
                {
                    a[i]='0';
                    digsum = res;
                    if(s==t)
                        s++;
                    else
                        t++;
                }
            }
        }
        //cout<<s<<t<<endl;
        if(s>t)
            printf("Case %d: S\n", ++caseno);
        else
            printf("Case %d: T\n", ++caseno);
    }
    return 0;
}

Please Help Me Out ..

[Zyaad Jaunnoo]

Why It is Getting WA?? I have Checked All The Tests of Samples , uDebug and OJ Borads . All are Ok . But I am getting WA .
.
.

Your code is almost correct.. But, you missed a "break" statement when picking a move.

INPUT

Code: Select all

1
2354456456587544678986756467889867

OUTPUT

Code: Select all

Case 1: S