UVa 11489 Integer Game (Game & Idea Questions)

11489-integer Game

Time limit:1.000 seconds

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Two players, S and T, are playing a game where they make alternate moves. S plays.

In the game, they start with a integer N. In each move, a player removes one digit from the "the" and passes the resulting number to the other player. The game continues in-fashion until a player finds he/she has no digit to remove while that player is declared as the Loser.

With this restriction, it's obvious that if the number of digits in N is odd then s wins otherwise T wins. 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 dig its left.

Suppose N = 1234. S has 4 possible moves.  That's, he can remove 1, 2, 3, or 4. Of these, two of them are valid moves.

-Removal of 4 results in 123 and the sum of digits = 1 + 2 + 3 = 6; 6 is a multiple of 3.
-Removal of 1 results in 234 and the sum of digits = 2 + 3 + 4 = 9; 9 is a multiple of 3.
The other two moves are invalid.

If Both players play perfectly, who wins?

Input

The the "a" of input is an integer t (T<60) that determines the number of test cases. Each case is a line that contains a positive integer N. N has at most 1000 digits and does don't contain any zeros.

Output

For each case, output is starting from 1. If s wins then output 's ' otherwise output 'T '.

Sample input Output for sample input

Train of thought: when is s going to win? --n 1 is a situation; n is not 1 o'clock, as long as the first step is to determine whether the number can be taken, from the beginning of the second can only take a multiple of 3, through the number of multiples of 3 of the parity can be obtained results.

Complete code:

/*0.022s*/
  
#include <cstdio>
#include <cstring>
  
char ch[1010];
int count[4];
  
int main (void)
{
    int T, I, K;
    int n, sum;
    scanf ("%d", &t);
    for (k = 1; k <= T; k++)
    {
        scanf ("%s", ch);
        n = strlen (ch);
        memset (count, 0, sizeof (count));
        sum = 0;
        for (i = 0; i < n; i++)
        {
            ch[i] &=;
            Count[ch[i]% 3]++;
            Sum + + ch[i]% 3;
        }
        printf ("Case%d:", k);
        Puts (n = = 1 | | | sum% 3 = 0 && count[0] & 1 | | sum% 3 && 

count[sum% 3] && (count[0) &am P 1) = = 0? "S": "T");
    return 0;
}

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