UVa 11489 - Integer Game （簡單博弈 腦筋急轉彎）

I	Integer Game

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

Problem Setter: Sohel Hafiz

Special Thanks: Shamim Hafiz, Md. Arifuzzaman Arif

題意：

給出一個數字串Ｎ，兩個人輪流從中取出一個數字，要求每次取完之後剩下的數之和是３的倍數（這裡之前看ｌｒｊ的描述了解錯了）。不能取者輸。問先手的輸赢

如果可以，第一次取走後變為了３的倍數，則以後取的一定是３的整數倍，直至取完所有是３的整數倍的數。

是以先統計MOD 3為0 1 2的數有多少，然後先手先取一個數使得其他數之和MOD 3 = 0，不能取則輸出Ｔ

然後輪到第二個人取數了，統計MOD ３==0的數有多少，如果是偶數，則先手勝，反之先手輸。

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 1000 + 20;
char str[maxn];
int cnt[3];

int main() {
    int T;

    scanf("%d", &T);
    for(int kase=1; kase<=T; kase++) {
        scanf("%s", str);
        int n = strlen(str);
        int sum = 0;
        cnt[0] = cnt[1] = cnt[2] = 0;
        for(int i=0; i<n; i++) {
            sum += str[i] - '0';
            cnt[(str[i]-'0')%3]++;
        }
        int t = sum % 3;
        cnt[t]--;
        printf("Case %d: ", kase);
        if(cnt[t] < 0) {
            puts("T");
            continue;
        }
        t = cnt[0];
        if(t&1) puts("T");
        else puts("S");
    }

    return 0;
}