Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Background

As you should know, a formal system consists of a set of axioms and set of inference or production rules. Theorems, in general, are results that can be obtained upon the axioms and the inference rules of the system. Deciding if a statement is a theorem of a given formal system is not a trivial matter; in fact, for some systems it is not even possible, as we know by Gödel's incompleteness theorem.

The Problem

ME is a formal system with an infinite number of axioms and just one production rule. There are only three symbols that may appear in any axiom of ME: the capital letters M and E, and the symbol ?. In spite of the infinite number of axioms, there is an easy way of identifying them:

• Axiom definition: An axiom of ME is a string of the form xM?Ex? where x is a string of one or more ? symbols.

Notice that the word xM?Ex? itself is not an axiom, because x is not a valid symbol of ME; it is just a pattern to describe the axioms. But ??M?E??? is indeed an axiom, because it fits the pattern, just replacing x with ??

There is another type of valid strings in the system ME: the theorems. Every axiom is a theorem of ME. In addition, the following production rule gives rise to an infinite number of theorems:

• Production rule: If xMyEz is a theorem then xMy?Ez? is also a theorem, where x,y,z are strings of one or more ? symbols.

For instance, the string ??M?E??? is a theorem, because it is an axiom, and the string ??M??E???? is also a theorem, since it can be obtained upon the theorem ??M?E??? and the production rule.

Can we decide if a given string is a theorem of ME? Sure!, there is a decision algorithm for the system ME. But your goal is less general: just write a program that reads strings with at most 50 symbols and decides if they are theorem of ME.

Input

The input begins with a single positive integer N, on a line by itself, indicating the number of input lines following. Each input line contains a string with 1 to 50 characters (without spaces), but not necessarily restricted to the symbols of ME.

Output

For each input line, the program must print an output line with the word "theorem" if the test line contains a theorem, or the word "no-theorem" in other case.

Sample Input

7
??M?E??? 
xM?Ex?
??M??E????
M?E?
??m?e???
?M?E??
12M12E????

Sample Output

theorem
no-theorem
theorem
no-theorem
no-theorem
theorem
no-theorem

---

我們先將 aMbEc，a, b, c 分別代表有多少個 '?'
基礎定義會得到 b = 1, a = c-1, a > 0
產生規則為 b++, c++

那麼直接計算 a, b, c 個數。
然後看能不能湊回基礎定義。

#include <stdio.h>
int check(char s[]) {
    int a, b, c, i = 0;
    a = 0, b = 0, c = 0;
    while(s[i] == '?')  a++, i++;
    if(s[i] != 'M') return 0;
    i++;
    while(s[i] == '?')  b++, i++;
    if(s[i] != 'E') return 0;
    i++;
    while(s[i] == '?')  c++, i++;
    if(s[i] != '\0')    return 0;
    return a == (c-(b-1))-1 && a && b;
}
int main() {
    int testcase;
    char s[105];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%s", s);
        puts(check(s) ? "theorem" : "no-theorem");
    }
    return 0;
}