The Description Problem
=== Op Tech Briefing, 2002/11/02 06:42 CST ===
"Locked in at The Item IS A Safe Klein (Klein safe) behind in Painting at The SECOND, A-Floor Library.
Klein Safes are extremely rare (rare);. most of them, along with Klein and his factory ( along with Klein and his function), were Destroyed in World War II
fortunately Old Brumbaugh (Brembo, names) from research knew Klein's Secrets and wrote Them Down the before of He died.
a Klein Safe has TWO Distinguishing features (features): a combination lock (combination lock , combination lock) that uses letters instead of numbers, and an engraved quotation ( deep deep into the quotations)
ON Door the.
a Klein and Quotation Always the contains BETWEEN Twelve Five DISTINCT uPPERCASE (capital letters) letters, usually at the beginning of sentences, and mentions ( referred to) one or more numbers.
Five of the uppercase letters form（形式，排列） the combination that opens the safe.
By combining the digits（数字） 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
Mid-Central USA 2002
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5 with reference to the function parameters, reference characters to be added after the variable type declaration
C language code as follows:
#include <stdio.h> #include <the iostream> #include < String > #include <algorithm> the using namespace STD; char A [ . 5 ]; int Solution ( Long Long , String ); void the BubbleSort ( String &); // with reference to the function parameter declaration int main () { int Long Long n-= 0 ; String STR = "" ; the while (CIN >> n-) { CIN>> STR; // COUT << STR; IF (n-== 0 && STR == " the END " ) BREAK ; the else the BubbleSort (STR); // this strings alphabetically descending order solution (n, str ); } return 0 ; } int Solution ( Long Long n-, String STR) { for ( int I = str.size () - . 1 ; I> = . 4 ; i-- ) for ( int J = I-1;j>=3;j--) for(int k=j-1;k>=2;k--) for(int l=k-1;l>=1;l--) for(int m=l-1;m>=0;m--) { a[0]=str[i]; a[1]=str[j]; a[2]=str[k]; a[3]=str[l]; a[4]=str[m]; do { if( (a[0]-64) - (long long)pow( (a[1]-64),2) + (long long)pow( (a[2]-64),3) - (long long)pow( (a[3]-64),4) + (long long)pow( (a[4]-64),5) ==n) { printf("%c%c%c%c%c\n",a[0],a[1],a[2],a[3],a[4]); return 1; } } while ( prev_permutation(a,a+5) ); //Get smaller than the current sequence sequence sequence dictionaries (dictionary is sequentially before a sequence) returns 0 if not, if there returns 1. } the printf ( " NO Solution \ n- " ); return 0 ; } void the BubbleSort ( String & STR) { char TEMP; for ( int I = 0 ; I <str.size () - . 1 ; ++ I) for ( int J = 0 ; J <str.size () - . 1 -i; ++ J) IF ( STR [J]> STR [J + . 1 ]) { temp =str [s]; str [s] = str [j + 1 ]; str [j + 1 ] = temp; } Return ; }