杭电-1015Safecracker（DFS）

电玩女神 2022-08-09 14:51 111阅读 0赞

# Safecracker #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)**

Total Submission(s): 10906 Accepted Submission(s): 5584

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

题目没读完，大致懂了啥意思，就是前面给个数，后面给个字符串，A对应1，B对应2......，依次类推，要你跳出5个字符使得 v - w^2 + x^3 - y^4 + z^5 = target

必须要遍历所有的情况，所以dfs不解释！

另外，字典序的意思就是strcmp(s1,s2)>0就取s1

代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #define INF 0x3f3f3f3f
    using namespace std;
    int t,len,used[100];
    char str[100],best[6],str1[100];
    int bes;
    bool is_solution(char s[])
    {
    		int r=s[0]-'A'+1-(s[1]-'A'+1)*(s[1]-'A'+1)+pow(s[2]-'A'+1,3)-pow(s[3]-'A'+1,4)+pow(s[4]-'A'+1,5);
    		if(r==t)
    		return true;
    		return false;
    }
    void dfs(int x)
    {
    	int i;
    	if(x>=5)
    	{
    		if(is_solution(str1)) 
    		{
    			if(strcmp(str1,best)>0)
    			{
    				for(i=0;i<5;i++)
    				best[i]=str1[i];
    			}
    		}
    		return ;
    	}
    	else{
    		for(i=0;i<len;i++)
    		{
    			if(!used[i])
    			{
    				str1[x]=str[i];
    				used[i]=1;
    				dfs(x+1);
    				used[i]=0;
    			}	
    		}
    	}
    }
    int main()
    {
    	while(scanf("%d%s",&t,str),t&&strcmp(str,"END\0"))
    	{
    		memset(used,0,sizeof(used));
    		memset(best,'0',sizeof(best));
    		len=strlen(str);
    		bes=INF;
    		dfs(0);
    		best[5]='\0';
    		if(best[0]=='0')
    		printf("no solution\n");
    		else
    		printf("%s\n",best);
    	}
    	return 0;
    }