UVA 11489 博弈思维题

http://vjudge.net/contest/view.action?cid=46225#problem/H

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

两个人取数，规则是每当一个人取过之后要么没有数了，要么剩下的数的和为3的倍数，两个人都采取最优的策略。

当n=1先手胜。n>1时，第一个人取候剩下的数是3的倍数，那么第二人去这能取走是三的倍数的一个数，否则无法保证剩下的数是三的倍数。那么，问题就转化成了三的倍数的数的数目。如果第一个人未取时恰好是三的倍数，那么，三的倍数的数个数为奇数就是后手胜，反之先手胜。若第一个人未取时不是三的倍数，那么，三的倍数的数个数为奇数就是先手胜，反之后手胜。

代码实现：

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;char a[10005];int main(){    int T,sum1,count1;    int tt=1;    char ans;    scanf("%d",&T);    while(T--)    {        scanf("%s",a);        int n=strlen(a);        int count=0;        int sum=0;        ans='T';        for(int i=0;i<n;i++)        {            count+=a[i]-'0';            if((a[i]-'0')%3==0)                sum++;        }        for(int i=0;i<n;i++)        {            count1=count-(a[i]-'0');//判断第一个取走哪一个数剩下的满足条件            sum1=sum;            if(count1%3==0)            {                if((a[i]-'0')%3==0)//第一个取走的是3的倍数                    sum1--;                if(sum1%2==0)                {                    ans='S';                    break;                }            }        }        printf("Case %d: %c\n",tt++,ans);    }    return 0;}