ZOJ 1403 解密

参考自：https://www.cnblogs.com/ECJTUACM-873284962/p/6412212.htmlSafecracker

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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=== 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

no solution

no solution

no solution

 

 

题意：

　　密码序列由一系列大写字母组成，在解密序列不唯一的情况下，按字典序输出最后一个，解密公式：v - w^2 + x^3 - y^4 + z^5 = target

 

　　由于题目中解的值域已经确定，解元素中的v,w,x,y,z都是题目中给定集合中的一个元素，数据范围较小枚举便可。

解题思路：

        由于题目中解的值域已经确定，解元素中的v,w,x,y,z都是题目中给定集合中的一个元素，数据范围较小枚举便可。

      *注意：由于题目求得是密码序列是按字典顺序的最后一个，所以再次我将之先降序排序，这样一来找到的第一个符合条件的肯定便是最后的！

 

代码

 

#include <bits/stdc++.h>
using namespace std;
char letters[15];
int value[15],target;
void process(int len)
{
    int a,b,c,d,e;
    for(a=0;a<len;a++)
        for(b=0;b<len;b++)
            if(a!=b)
                for(c=0;c<len;c++)
                    if(a!=c&&b!=c)
                        for(d=0;d<len;d++)
                            if(a!=d&&b!=d&&c!=d)
                                for(e=0;e<len;e++)
                                    if(a!=e&&b!=e&&c!=e&&d!=e)
                                        if(value[a]-pow(value[b],2.0)+pow(value[c],3.0)-pow(value[d],4.0)+pow(value[e],5.0)==target)
                                        {
                                            printf("%c%c%c%c%c\n",value[a]+'A'-1,value[b]+'A'-1,value[c]+'A'-1,value[d]+'A'-1,value[e]+'A'-1);
                                            return;
                                        }
                                         printf("no solution\n");
                                    }
bool compare(int a,int b)
{
    return a>b;
}
int main()
{
    int i;
    while(scanf("%d%s",&target,letters)!=EOF)
    {
        if(target==0&&strcmp(letters,"END")==0)
            return 0;
        i=0;
        while(letters[i])
        {
            value[i]=letters[i]-'A'+1;
            i++;
        }
        sort(value,value+i,compare);
        process(i);
    }
    return 0;
}

 

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