4月 04, 2009

【解題】f91

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ACM Volume CVI 10696 - f91


Background

McCarthy is a famous theorician of computer science. In his work, he defined a recursive function, called f91, that takes as input a positive integer N and returns a positive integer defined as follows:
  • If N ≤ 100, then f91(N) = f91(f91(N+11));
  • If N ≥ 101, then f91(N) = N-10.

The Problem

Write a program, that computes McCarthy's f91.


Input

The input tests will consist of a series of positive integers, each integer is at most 1,000,000. There will be at most 250,000 test cases. Each number is on a line on its own. The end of the input is reached when the number 0 is met. The number 0 shall not be considered as part of the test set.


Output

The program shall output each result on a line by its own, following the format given in the sample output.


Sample Input

500
91
0


Sample Output

f91(500) = 490
f91(91) = 91


解題思考

  這一題並不難,直接照著題目定義,用遞迴寫也可以 AC。

  不過若是仔細分析下來,其實可以發現:當 N 小於等於 100 時,f91(N) 必等於 91。

  總而言之,這題其實連遞迴都不需要用,只需要一個 if 條件判別就行了。


參考解答(C++)

#include <iostream>

using namespace std;

int main(void)
{
    while (1)
    {
        unsigned int n;
        cin >> n;
        if (n == 0) { break; }

        if (n <= 100)
        {
            // 若 n 小於 100, f91(n) 必等於 91
            cout << "f91(" << n << ") = 91" << endl;
        }
        else
        {
            cout << "f91(" << n << ") = " << n - 10 << endl;
        }
    }

#ifndef ONLINE_JUDGE
    system("pause");
#endif

    return 0;
}

3 回覆:

yen3 提到...

這題當初寫的時候,簡單到我不知道怎麼說了XD

小參 提到...

是沒錯
不過這題還比不上 11172 那題那麼簡單XD

匿名 提到...

若沒想到<100的都是91
就不簡單了

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