HDU-1015-Safecracker(Java && 暴力大法好 && 簡單字典序)

標籤：hdu   java   字典序   簡單題   acm   

Safecracker Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9965    Accepted Submission(s): 5097


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 ABCDEFGHIJKL11700519 ZAYEXIWOVU3072997 SOUGHT1234567 THEQUICKFROG0 END

 
Sample Output

LKEBAYOXUZGHOSTno solution

 
SourceMid-Central USA 2002 
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看著這麼長的英文,亞曆山大！！但是它說了這麼長要表達的意思其實很簡單！就是給定一個目標數字target,再給你一個備選字串(5~12個字元)，要你從這個備選字串（字串全都有大寫字母構成）中選5個字元出來，滿足題目給出的等式——v - w^2 + x^3 - y^4 + z^5 = target （v,w,x,y,z分別表示這個選取的5個字元的數字序號，‘A’=1,‘B=2‘依次類推）
並且選擇的這5個字串必須是所有可能情況中按照字典序最大的情況
簡單分析下就可以看出，就是一個組合問題，問題解的最大規模就是12取5，就是12*11*10*9*8*7，而最小規模是5取5，所以應該用枚舉法就可以搞定。不過，枚舉之前先排個序，就可以保證輸出的是符合要求的最大的那個了

//暴力水過！簡單的字典序問題弄了菜鳥博主半個多小時！import java.io.*;import java.util.*;public class Main{public static void main(String[] args){// TODO Auto-generated method stubScanner input = new Scanner(System.in);while (input.hasNext()){int target = input.nextInt();boolean flag = false;                                      String str = input.nextLine();                               str = str.replaceAll(" ", "");                            //去掉字串前面的空格if (target == 0 && str.compareTo("END") == 0)             //如果滿足條件中止程式break;char c[] = str.toCharArray();                             //字串轉換為字元數組int a[] = new int[str.length()];for (int i = 0; i < c.length; i++){a[i] = CMP(c[i]);                                    //字元數群組轉換為數字數組}Arrays.sort(a);                                          //快速排序方法，但是此時數組順序是從小到大for (int i = c.length - 1; i >= 0; i--){for (int j = c.length - 1; j >= 0; j--){if (j == i)                                    continue;for (int k = c.length - 1; k >= 0; k--){if (k == i || k == j)continue;for (int l = c.length - 1; l >= 0; l--){if (l == i || l == j || l == k)continue;for (int m = c.length - 1; m >= 0; m--){if (m == i || m == j || m == k || m == l)continue;if ((a[i] - (a[j] * a[j])                                    //代入公式求解+ (a[k] * a[k] * a[k])- (a[l] * a[l] * a[l] * a[l]) + (a[m]* a[m] * a[m] * a[m] * a[m])) == target){flag = true;char c1 = (char) (a[i] + 'A' - 1);char c2 = (char) (a[j] + 'A' - 1);char c3 = (char) (a[k] + 'A' - 1);char c4 = (char) (a[l] + 'A' - 1);char c5 = (char) (a[m] + 'A' - 1);System.out.println(c1 + "" + c2 + "" + c3 + "" + c4 + "" + c5);break;}}if (flag)break;}if (flag)break;}if (flag)break;}if (flag)break;}if (!flag)System.out.println("no solution");}}public static int CMP(char c){return ((c - 'A') + 1);                                                  //A字元代表的數字記號是1}}

 

HDU-1015-Safecracker(Java && 暴力大法好 && 簡單字典序)