UVa 11489 Integer Game （博弈&想法题）

UVa 11489 Integer Game （博弈&想法题）2014-04-26 csdn博客 synapse711489 - Integer Game

Time limit: 1.000 seconds

http://uva.onlinejudge.org/index.php？ option=com_onlinejudge&Itemid=8&category=467&page=show_problem&problem=24 84

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

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;}