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

Problem C: Cantor

The ternary expansion of a number is that number written in base 3. A number can have more than one ternary expansion. A ternary expansion is indicated with a subscript 3. For example, 1 = 1₃ = 0.222...₃, and 0.875 = 0.212121...₃.

The Cantor set is defined as the real numbers between 0 and 1 inclusive that have a ternary expansion that does not contain a 1. If a number has more than one ternary expansion, it is enough for a single one to not contain a 1.

For example, 0 = 0.000...₃ and 1 = 0.222...₃, so they are in the Cantor set. But 0.875 = 0.212121...₃ and this is its only ternary expansion, so it is not in the Cantor set.

Your task is to determine whether a given number is in the Cantor set.

Input Specification

The input consists of several test cases.

Each test case consists of a single line containing a number x written in decimal notation, with 0 <= x <= 1, and having at most 6 digits after the decimal point.

The last line of input is END. This is not a test case.

Sample Input

0
1
0.875
END

Output Specification

For each test case, output MEMBER if x is in the Cantor set, and NON-MEMBER if x is not in the Cantor set.

Output for Sample Input

MEMBER
MEMBER
NON-MEMBER

---

Malcolm Sharpe

題目要求三進制的浮點數表示法。並且判斷他是不是一個 Cantor set 的成員。
Cantor set 定義是不會出現 1 的數字。

根據二進制找浮點數表示法的方式，一直乘 3 就可以轉換了。
檢查到循環為止。


#include <stdio.h>
#include <set>
using namespace std;
int main() {
    char s[105];
    while(scanf("%s", s) == 1) {
        if(s[0] == 'E') break;
        int digit = 0, i;
        int a = 0, b = 0, c = 1;
        for(i = 0; s[i]; i++) {
            if(s[i] == '.') {
                for(i++; s[i]; i++, digit++)
                    b = b*10 + s[i]-'0', c = c*10;
                break;
            } else {
                a = a*10 + s[i]-'0';
            }
        }
        if((a == 1 && b == 0) || (a == 0 && b == 0)) {
            puts("MEMBER");
            continue;
        }
        set<int> R;
        int NON = 0;
        while(!NON) {
            if(b*3/c == 1)
                NON = 1;
            b = (b*3)%c;
            if(R.find(b) != R.end())
                break;
            R.insert(b);
        }
        if(!NON)
            puts("MEMBER");
        else
            puts("NON-MEMBER");
    }
    return 0;
}