UVA 11489 Integer Game （博弈）

1227: Integer Game

Time Limit: 1 Sec Memory Limit:128 MB

Submit: 9 Solved: 4

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Board]

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;
}