UVa 11489 Integer Game (博弈&想法题)

11489 - Integer Game

Time limit: 1.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=467&page=show_problem&problem=2484

Two players, S and T, are playing a game where they make alternate moves. S plays first. 
In this game, they start with an integer N. In each move, a player removes one digit from the integer and passes the resulting number to the other player. The game continues in this fashion until a player finds he/she has no digit to remove when 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 digits left.

Suppose N = 1234. S has 4 possible moves. That is, 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 first line 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 not contain any zeros.

Output
For each case, output the case number starting from 1. If S wins then output ‘S’ otherwise output ‘T’.

Sample Input      Output for Sample Input

3 4 33 771

Case 1: S Case 2: T Case 3: T

思路：S何时会赢？——n为1是一种情况；n不为1时，只要判断第一步能否取数即可，从第二步开始就只能取3的倍数了，通过3的倍数的个数的奇偶性就可求得结果。

完整代码：

/*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] &= 15;
			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] & 1) == 0 ? "S" : "T");
	}
	return 0;
}