HDOJ1015.Safecracker

試題請參見: http://acm.hdu.edu.cn/showproblem.php?pid=1015

題目概述

=== Op tech briefing, 2002/11/02 06:42 CST ===

“The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in World War II. Fortunately old Brumbaugh from research knew Klein’s secrets and wrote them down before he died. A Klein safe has two distinguishing features: a combination lock that uses letters instead of numbers, and an engraved quotation on the door. A Klein quotation always contains between five and twelve distinct uppercase letters, usually at the beginning of sentences, and mentions one or more numbers. Five of the uppercase letters form the combination that opens the safe. By combining the digits from all the numbers in the appropriate way you get a numeric target. (The details of constructing the target number are classified.) To find the combination you must select five letters v, w, x, y, and z that satisfy the following equation, where each letter is replaced by its ordinal position in the alphabet (A=1, B=2, …, Z=26). The combination is then vwxyz. If there is more than one solution then the combination is the one that is lexicographically greatest, i.e., the one that would appear last in a dictionary.”

v - w^2 + x^3 - y^4 + z^5 = target

“For example, given target 1 and letter set ABCDEFGHIJKL, one possible solution is FIECB, since 6 - 9^2 + 5^3 - 3^4 + 2^5 = 1. There are actually several solutions in this case, and the combination turns out to be LKEBA. Klein thought it was safe to encode the combination within the engraving, because it could take months of effort to try all the possibilities even if you knew the secret. But of course computers didn’t exist then.”

=== Op tech directive, computer division, 2002/11/02 12:30 CST ===

“Develop a program to find Klein combinations in preparation for field deployment. Use standard test methodology as per departmental regulations. Input consists of one or more lines containing a positive integer target less than twelve million, a space, then at least five and at most twelve distinct uppercase letters. The last line will contain a target of zero and the letters END; this signals the end of the input. For each line output the Klein combination, break ties with lexicographic order, or ‘no solution’ if there is no correct combination. Use the exact format shown below.”

題意是說将ABCD…Z映射成1, 2, 3, …, 26. 給出一個目标數值(target), 和一組字元序列. 問該序列中的字母(對應的數值分别為v, w, x, y, z)能否選取5個數使得

v - w^2 + x^3 - y^4 + z^5 = target

.

對了, 若有多組符合條件的v, w, x, y, z, 輸出字典序最後的一組.

解題思路

感覺暴力搜尋就可以, 是以寫DFS就沒有太大的意義. 試了一下果然不逾時.

既然要求輸出字典序最後的一組, 那就直接倒過來搜尋就可以了. :)

源代碼

#include <iostream>
#include <string>
#include <sstream>
#include <algorithm>

bool compareTo(const char& a, const char& b) {
    if( a > b ) {
        return true;
    } else {
        return false;
    }
}

std::string getSolution(int target, const std::string& alphabet) {
    for ( int i = 0; i < alphabet.size(); ++ i ) {
        for ( int j = 0; j < alphabet.size(); ++ j ) {
            for ( int k = 0; k < alphabet.size(); ++ k ) {
                for ( int m = 0; m < alphabet.size(); ++ m ) {
                    for ( int n = 0; n < alphabet.size(); ++ n ) {
                        if ( i != j && i != k && i !=m && i != n && j !=k && j != m && j != n && k != m && k != n && m !=n ) {
                            int v = alphabet[i] - 'A' + 1;
                            int w = alphabet[j] - 'A' + 1;
                            int x = alphabet[k] - 'A' + 1;
                            int y = alphabet[m] - 'A' + 1;
                            int z = alphabet[n] - 'A' + 1;

                            if ( v - w * w + x * x * x - y * y * y * y + z * z * z * z * z == target ) {
                                std::ostringstream ss;
                                ss << alphabet[i] << alphabet[j] << alphabet[k] << alphabet[m] << alphabet[n];
                                return ss.str();
                            }
                        }
                    }
                }
            }
        }
    }
    return "no solution";
}

int main() {
    int target = 0;
    std::string alphabet;

    while ( std::cin >> target >> alphabet ) {
        if ( target == 0 && alphabet == "END" ) {
            break;
        }
        std::sort(alphabet.begin(), alphabet.end(), compareTo);
        std::cout << getSolution(target, alphabet) << std::endl;
    }
    return 0;
}