UVa 11489 - Integer Game （简单博弈 脑筋急转弯）

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

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

题意：

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之和是３的倍数（这里之前看ｌｒｊ的描述理解错了）。不能取者输。问先手的输赢

如果可以，第一次取走后变为了３的倍数，则以后取的一定是３的整数倍，直至取完所有是３的整数倍的数。

所以先统计MOD 3为0 1 2的数有多少，然后先手先取一个数使得其他数之和MOD 3 = 0，不能取则输出Ｔ

然后轮到第二个人取数了，统计MOD ３==0的数有多少，如果是偶数，则先手胜，反之先手输。

#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 WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef 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;}