BNU19907 UVA11489 Integer Game

Integer Game

Time Limit: 1000ms
Memory Limit: 131072KB
This problem will be judged on UVA. Original ID: 11489

64-bit integer IO format: %lld      Java class name: Main
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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, 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 Output for Sample Input

3 4 33 771

Case 1: S Case 2: T Case 3: T

https://www.bnuoj.com/v3/problem_show.php?pid=19907

给出一个数字串N,两个人轮流从中取出一个数字,要求每次取完之后剩下的数是3的倍数,不能取着输.如果两个游戏者都足够聪明,谁会取胜,输入非空数字串N(N由不超过1000个非0的数字组成),如果先手胜,输出S,否则输出T.

#include<bits/stdc++.h>
using namespace std;
int main()
{
char str[1010];
int t,h,sum,flag,k,i,g;
scanf("%d",&t);
for(h=1; h<=t; h++)
{
scanf("%s",str);
sum=0;
k=0;
for(i=0; str[i]!=0; i++)
{
sum+=str[i]-48;
if((str[i]-48)%3==0)
k++;
}
printf("Case %d: ",h);
if(sum%3!=0)
{
g=sum%3;
flag=0;
for(i=0; str[i]!=0; i++)
if((str[i]-48)%3==g)
flag=1;
if(!flag)
{
printf("T\n");
continue;
}
if(k%2==0)
{
printf("S\n");
continue;
}
else
{
printf("T\n");
continue;
}
}
else
{
if(k&1)
{
printf("S\n");
continue;
}
else
{
printf("T\n");
continue;
}
}
}
return 0;
}