Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 3277 | Accepted: 1741 |
Description
"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 - w2 + x3 - y4 + z5 = target
"For example, given target 1 and letter set ABCDEFGHIJKL, one possible solution is FIECB, since 6 - 92 + 53 - 34 + 25 = 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."
"Develop a program to find Klein combinations in preparation for field deployment. Use standard test methodology as per departmental regulations.
v - w2 + x3 - y4 + z5 = target
"For example, given target 1 and letter set ABCDEFGHIJKL, one possible solution is FIECB, since 6 - 92 + 53 - 34 + 25 = 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."
"Develop a program to find Klein combinations in preparation for field deployment. Use standard test methodology as per departmental regulations.
Input
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.
Output
For each line output the unique Klein combination, 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
Regionals 2002 >> North America - Mid-Central USA
问题链接:UVALive2536 POJ1248 HDU1015 ZOJ1403 Safecracker
问题简述:给一个密码值,给一个字符串,找出6个字符使得经过指定的运算之后等于密码值,输出这些字符。
问题分析:这个问题用暴力法来解。
程序说明:(略)
题记:(略)
参考链接:(略)
AC的C++语言程序如下:
/* UVALive2536 POJ1248 HDU1015 ZOJ1403 Safecracker */ #include <iostream> #include <algorithm> #include <string> #include <stdio.h> using namespace std; bool cmp(char a,char b) { return a > b; } int calc(int v,int w,int x,int y,int z) { return v - w*w + x*x*x - y*y*y*y + z*z*z*z*z; } void solve(int target, string& s) { int n = s.length(); for(int z=0; z<n; z++) { for(int y=0; y<n; y++) { if(y == z) continue; for(int x=0; x<n; x++) { if(x == z || x == y) continue; for(int w=0; w<n; w++) { if(w == z || w == y || w == x) continue; for(int v=0; v<n; v++) { if(v == z || v == y || v == x || v == w) continue; if(calc(s[z]-'A'+1, s[y]-'A'+1, s[x]-'A'+1, s[w]-'A'+1, s[v]-'A'+1) == target) { printf("%c%c%c%c%c ", s[z], s[y], s[x], s[w], s[v]); return ; } } } } } } printf("no solution "); } int main() { int target; string s; while(cin >> target >> s && target) { sort(s.begin(), s.end(), cmp); solve(target, s); } return 0; }