• HDU 1015 Safecracker 解决问题的方法


    Problem Description
    === 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."
     

    Sample Input
    1 ABCDEFGHIJKL 11700519 ZAYEXIWOVU 3072997 SOUGHT 1234567 THEQUICKFROG 0 END
     

    Sample Output
    LKEBA YOXUZ GHOST no solution
     

    Source


    水题一枚。呵呵,纯粹暴力法过的。

    使用C++比C能够大大降低出错的可能哦。

    注意:要求按最大的字典序输出。故此要排序。


    奉献给个小技巧:

    sort(str.rbegin(), str.rend());

    这个rbegin是逆序的iterator了,故此这样排序处理就是逆序排序的字典序了。


    #include <iostream>
    #include <stdio.h>
    #include <string>
    #include <algorithm>
    using namespace std;
    
    __int64 num;
    string str;
    __int64 a, b, c, d, e;
    
    inline __int64 cal()
    {
    	return a - b*b + c*c*c - d*d*d*d + e*e*e*e*e;
    }
    
    bool getCombination()
    {
    	for (int i = 0; i < (int)str.size(); i++)
    	{
    		a = str[i] - 'A' + 1;
    		for (int j = 0; j < (int)str.size(); j++)
    		{
    			if (j == i) continue;
    			b = str[j] - 'A' + 1;
    			for (int k = 0; k < (int)str.size(); k++)
    			{
    				if (k == i || k == j) continue;
    				c = str[k] - 'A' + 1;
    				for (int u = 0; u < (int)str.size(); u++)
    				{
    					if (u == k || u == i || u ==j) continue;
    					d = str[u] - 'A' + 1;
    					for (int v = 0; v < (int)str.size(); v++)
    					{
    						if (v == u || v == k || v == i || v == j) continue;
    						e = str[v] - 'A' + 1;
    						if (cal() == num) return true;
    					}
    				}
    			}
    		}
    	}
    	return false;
    }
    
    int main()
    {
    	while (cin>>num && cin>>str)
    	{
    		if (num == 0LL && str == "END") break;
    		sort(str.rbegin(), str.rend());
    		if (getCombination())
    			cout<<char(a+'A'-1)<<char(b+'A'-1)<<char(c+'A'-1)<<char(d+'A'-1)
    			<<char(e+'A'-1)<<endl;
    		else cout<<"no solution
    ";
    	}
    	return 0;
    }



    版权声明:笔者靖心脏,景空间地址:http://blog.csdn.net/kenden23/,只有经过作者同意转载。

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  • 原文地址:https://www.cnblogs.com/hrhguanli/p/4869832.html
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