hdu1015Safecracker

最新推荐文章于 2021-05-31 18:00:25 发布

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最新推荐文章于 2021-05-31 18:00:25 发布

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分类专栏： ACM搜索 ACM生涯文章标签： acm hdu1015 dfs 深搜

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Safecracker

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
 
 
题目意思：给出一个整数target，叫你利用给出的一串字符串中挑选5个数，满足v - w^2 + x^3 - y^4 + z^5 = target；
但是由于可能不止一种答案，所以输出字典序最大的答案（用strcmp即可，它是按照字典树排序的额）,用一个全局变量max[20]数组来记录，刚开始将其初始化，每当出现满足条件的答案，将其与max比较，如果大就更新max;
 
初一看暴力貌似可以解决，试了一下，感觉代码太麻烦，于是改用dfs,搜索起来还是很简单的,虽然出了点小意外,hahaha;
#include<stdio.h>
#include<string.h>
char max[20];
char n[20];
char str[100010];
int visit[100010],num[20],sum,t,len;
//visit标记  n数组存输入的，num数组用来转化的，sum表示算式左边答案 t代表右边答案 len代表字符串长度
void dfs(int count)			//count代表次数
{
	int i;
	if(count==5)	
	{
		n[5]='\0';
		for(i=0;i<5;i++)
			num[i]=n[i]-'A'+1;
		sum=num[0]-num[1]*num[1]+num[2]*num[2]*num[2]-num[3]*num[3]*num[3]*num[3]+num[4]*num[4]*num[4]*num[4]*num[4];
		if(sum==t)
		{
			if(strcmp(max,n)<0)			
				strcpy(max,n);			//更新max
		}
		return ;
	}
	for(i=0;i<len;i++)
	{
		if(!visit[i])
		{
			n[count]=str[i];
			visit[i]=1;
			dfs(count+1);
			visit[i]=0;
		}
	}
}

int main()
{
	int i;
	while(scanf("%d%s",&t,str)!=EOF)
	{
		strcpy(max,"#");
		if(t==0&&strcmp(str,"END")==0)
			break;
		len=strlen(str);
		for(i=0;i<len;i++)
		{
			memset(visit,0,sizeof(visit));
			visit[i]=1;
			n[0]=str[i];
			dfs(1);
			visit[i]=0;
		}
		if(strcmp(max,"#")==0)
			printf("no solution\n");
		else
			printf("%s\n",max);
	}
	return 0;
}
 
 
后来帮队友找错误，他用的是枚举，真烦，代码太烦了，看得头都大了，不过幸好还是找出来了，附上AC枚举代码
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int a[26];
char b[26];
void fun(char *s,int len)
{
    int i;
	for(i=0;i<len;i++)
		a[i]=(s[i]-'A')+1;
}

int cmp(const void *a,const void *b)
{
	return (*(char *)b - *(char *)a);
}
int main()
{
	int k=0;
	int sum,i,num;
	int q1,q2,q3,q4,q5;
	while(scanf("%d %s",&num,b)&&num!=0&&strcmp(b,"END")!=0)
	{
		
		bool flag=false;
		char c[5]={'0'};
		sum=0;
		getchar();
		int n=strlen(b);
		qsort(b,n,sizeof(b[0]),cmp);
		fun(b,n);
		for(q1=0;q1<n;q1++)
		{	
			for(q2=0;q2<n;q2++)
			{
				if(q2==q1)
				continue;
				for(q3=0;q3<n;q3++)
				{
					if(q3==q2||q3==q1)
					continue;
					for(q4=0;q4<n;q4++)
					{
						if(q4==q3||q4==q2||q4==q1)
						continue;
						for(q5=0;q5<n;q5++)
						{
							if(q5==q4||q5==q3||q5==q2||q5==q1)
							continue;
							sum=a[q1]-a[q2]*a[q2]+a[q3]*a[q3]*a[q3]-a[q4]*a[q4]*a[q4]*a[q4]+a[q5]*a[q5]*a[q5]*a[q5]*a[q5];
							if(sum==num)
							{
								flag=true;
								//strcpy(c,b);
								break;
							
							}
						}
						if(flag==true)
							break;
					}
					if(flag==true)
							break;
				}
				if(flag==true)
							break;
			}
			if(flag==true)
							break;
		}
		if(flag)
			printf("%c%c%c%c%c\n",a[q1]+'A'-1,a[q2]+'A'-1,a[q3]+'A'-1,a[q4]+'A'-1,a[q5]+'A'-1);
		else
			printf("no solution\n");
	}
	return 0;	
}



有问题欢迎大家指正!不胜感激