11489 Integer Game
Two players, S and T, are playing a game where they make alternate moves. S plays rst.
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 nds he/she has no digit to remove when that player is declared as the loser.
With this restriction, its 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 rst 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
3
4
33
771
Sample Output
Case 1: S
Case 2: T
Case 3: T
题意：
给出一字符串N，两人轮流从中取出一数字，要求每次取完之后剩下的数是3的倍数，不能取者输。N由不超过1000个非零数字组成。如果先手胜则输出S否则输出N。
分析：
要想取走一个数字之后剩下的数能被3整除，我们只需要统计N的每一个数字，其中能被3整除的数字有cnt个，数字之和为sum。那么我们只需要遍历每一个数字模拟先手取数字的情形，如果取走某个数字之后剩余的sum被3整除，则后手只能取走是3的倍数的数字，否则后手将会失败。如果先手无论取走哪一个数字都无法使得剩余的sum值被3整除则先手失败。
1 #include <cstdio>
2 #include <cstring>
3 #define MAX_LEN 1000
4 char str[MAX_LEN + 1];
5 int main(){
6 int T; scanf("%d",&T);
7 for(int kase = 1 ; kase <= T ; kase++){
8 scanf("%s",str);
9 char ans = ‘T‘;
10 printf("Case %d: ",kase);
11 int cnt = 0,sum = 0,len = strlen(str);
12 for(int i = 0 ; i < len ; i++){
13 sum += str[i] - ‘0‘;
14 if((str[i] - ‘0‘) % 3 == 0) cnt++;
15 }
16 for(int i = 0 ; i < len ; i++){
17 int tmp = sum - (str[i] - ‘0‘),tmpcnt = cnt;
18 if(tmp % 3 == 0){
19 if((str[i] - ‘0‘) % 3 == 0) tmpcnt = cnt - 1;
20 if(!(tmpcnt & 1)){
21 ans = ‘S‘;
22 break;
23 }
24 }
25 }
26 printf("%c\n",ans);
27 }
28 return 0;
29 }