Time Limit: 1 Sec Memory Limit:128 MB
Submit: 9 Solved: 4
Description
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, 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 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
3 4 33 771
Sample Output
Case 1: S Case 2: T Case 3: T
解析：
1.总和不为3的倍数时，判断第一个人能否取。
（1）能取，判断有多少个数为三的倍数即可。
（2）不能取， 输出T。
2.总和为3的倍数时，直接判断判断有多少个数为三的倍数即可。
AC代码；
#include <bits/stdc++.h>
using namespace std;
char s[1002];
bool vis[4];
int main(){
#ifdef sxk
freopen("in.txt", "r", stdin);
#endif // sxk
int T;
scanf("%d", &T);
for(int t=1; t<=T; t++){
memset(vis, false, sizeof(vis));
printf("Case %d: ", t);
scanf("%s", s);
int len = strlen(s);
int cnt = 0, sum = 0;
for(int i=0; i<len; i++){
int foo = s[i] - '0';
sum += foo;
if(foo % 3 == 0) cnt ++;
vis[foo % 3] = true;
}
if(sum % 3){
if(vis[sum % 3]) puts(cnt % 2 ? "T" : "S");
else puts("T");
}
else puts(cnt % 2 ? "S" : "T");
}
return 0;
}