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

Problem Setter: Sohel Hafiz
Special Thanks: Shamim Hafiz, Md. Arifuzzaman Arif

感觉题目意思有歧义。。。看了样例才明白的！

题目意思就是S先手取数  是剩下的数满足能被3整除。。。接着两人轮流取 到不能取为止

不能取的人输

因为 3  ，6 ，9这些数对3求余是等价的

1，4，7也一样

所以取1或4不会对最后结果造成影响！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
string st;
bool vis[1010];
int sum,cnt;
void init(){
    memset(vis,0,sizeof vis);
    sum = 0;
    cnt = st.size();
}
int main(){
    int ncase,T=1;
    cin >> ncase;
    while(ncase--){
        cin >> st;
        init();
        for(int i = 0; i < st.size(); i++){
            sum = (sum+st[i]-‘0‘)%3;
        }
        //cout<<sum<<endl;
        int turn = 0;
        if(sum!=0){
            for(int i = 0; i < st.size(); i++){
                if((st[i]-‘0‘)%3==sum){
                    vis[i] = 1;
                    //cout<<st[i]<<endl;
                    cnt--;
                    turn = 1;
                    break;
                }
            }
        }
        while(1){
            bool flag = 1;
            for(int i = 0; i < st.size(); i++){
                if(!vis[i]&&(st[i]-‘0‘)%3==0){
                    cnt--;
                    vis[i] = 1;
                    flag = 0;
                    if(turn) turn = 0;
                    else turn = 1;
                }
            }
            if(flag) break;
        }
        printf("Case %d: ",T++);
        if(turn==1) cout<<"S"<<endl;
        else cout<<"T"<<endl;
    }
    return 0;
}

UVa11489 - Integer Game,布布扣,bubuko.com

UVa11489 - Integer Game

原文：http://blog.csdn.net/mowayao/article/details/21866539