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  • HDOJ 1015 Safecracker

    Safecracker

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 15066    Accepted Submission(s): 7946


    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
     
    题意描述:
    给出一个目标值和一串仅包含大写字母的长度在5~12的字符串,
    字符串中A的值为1,B的值为2,,,Z为26,
    要求在这个字符串中找出5个不同的字符满足v-w*w+x*x*x-y*y*y*y+z*z*z*z*z
    等于目标值的同时,求出字典序最大的这样5个字符组成的字符串。

    Your idea:
    1.一开始你竟然会想到用5-12选5得组合算法,
    老铁,时间不够就算了,这里还要字典序输出 ,就算你先排序,但是还有重复的怎么办?
    2. 暴力枚举,不用在给定的字符串中找,直接从A-Z(不过要根据给定的字符串标记能用的),
    而且最先找到的就是最大字典序的,也不要用排序。
    3. 字符到数字,其实不用真的把字符转换成数字,直接从1-26枚举,最后在输出时转换成字符就行了。
    4.补充:int ('ascii') ==ascii,也就是说char(7)并不是对应'7';

    //COPY
    
    
    #include<cstdio>
    #include<cstring>
    int f[30],target;
    int main(){
        char s[30];
        while(scanf("%d%s",&target,s)!=EOF&&target){
            memset(f,0,sizeof(f));
            if(strlen(s)<5){
                printf("no solution");
                break;
            }
            bool key=false;
            for(int i=0;i<strlen(s);i++)
                f[s[i]-'@']=1;
            for(int a=26;a>0&&!key;a--)
            for(int b=26;b>0&&!key;b--){
                if(b!=a)
                for(int c=26;c>0&&!key;c--){
                    if(c!=b&&c!=a)
                    for(int d=26;d>0&&!key;d--){
                        if(d!=c&&d!=b&&d!=a)
                        for(int e=26;e>0&&!key;e--){
                            if(e!=d&&e!=c&&e!=b&&e!=a)
                            if(f[a]&&f[b]&&f[c]&&f[d]&&f[e])
                            if(a-b*b+c*c*c-d*d*d*d+e*e*e*e*e==target){
                                key=true;
                                printf("%c%c%c%c%c
    ",(char)a+64,(char)b+64,(char)c+64,(char)d+64,(char)e+64);
                            }
                        }
                    }
                }
            }
            if(!key) printf("no solution
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/biggan/p/7435101.html
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