Problem Description
The Tyrell corporation uses a state-of-the-art electronic document system that controls all aspects of document creation, viewing, editing, and distribution. Document security is handled via access control lists (ACLs). An ACL defines a set of entities that have access to the document, and for each entity defines the set of rights that it has. Entities are denoted by uppercase letters; an entity might be a single individual or an entire division. Rights are denoted by lowercase letters; examples of rights are a for append, d for delete, e for edit, and r for read.
The ACL for a document is stored along with that document, but there is also a separate ACL log stored on a separate log server. All documents start with an empty ACL, which grants no rights to anyone. Every time the ACL for a document is changed, a new entry is written to the log. An entry is of the form ExR, where E is a nonempty set of entities, R is a nonempty set of rights, and x is either "+", "–", or "=". Entry E+R says to grant all the rights in R to all the entities in E, entry E–R says to remove all the rights in R from all the entities in E, and entry E=R says that all the entities in E have exactly the rights in R and no others. An entry might be redundant in the sense that it grants an entity a right it already has and/or denies an entity a right that it doesn't have. A log is simply a list of entries separated by commas, ordered chronologically from oldest to most recent. Entries are cumulative, with newer entries taking precedence over older entries if there is a conflict.
Periodically the Tyrell corporation will run a security check by using the logs to compute the current ACL for each document and then comparing it with the ACL actually stored with the document. A mismatch indicates a security breach. Your job is to write a program that, given an ACL log, computes the current ACL.
Input
The input consists of one or more ACL logs, each 3–79 characters long and on a line by itself, followed by a line containing only "#" that signals the end of the input. Logs will be in the format defined above and will not contain any whitespace.
Output
For each log, output a single line containing the log number (logs are numbered sequentially starting with one), then a colon, then the current ACL in the format shown below. Note that (1) spaces do not appear in the output; (2) entities are listed in alphabetical order; (3) the rights for an entity are listed in alphabetical order; (4) entities with no current rights are not listed (even if they appeared in a log entry), so it's possible that an ACL will be empty; and (5) if two or more consecutive entities have exactly the same rights, those rights are only output once, after the list of entities.
Sample Input
MC-p,SC+c
YB=rde,B-dq,AYM+e
GQ+tju,GH-ju,AQ-z,Q=t,QG-t
JBL=fwa,H+wf,LD-fz,BJ-a,P=aw
#
Sample Output
1:CSc
2:AeBerMeYder
3:
4:BHJfwLPaw
https://blog.csdn.net/u012860063/article/details/40626077
#include<stdio.h>
#include<string.h>
int s[26][27];
int main()
{
char str[80];
int cas = 1, i, j, k;
while(gets(str) != NULL)//one test
{
if(str[0] == '#')
break;
printf("%d:", cas++);
int l = strlen(str);
memset(s, 0, sizeof(s));
for( i = 0; i < l; ++i)
{
for(j = i; str[j] != ',' && str[j] !='\0'; ++j)//get ACL log
{
if(str[j] == '=' || str[j] == '-' || str[j] == '+')
k = j;
}
int id1, id2;
for(int n = i; n < k; ++n)//up
{
id1 = str[n] - 'A';
if(str[k] == '=')
{
for(int m = 0; m < 27; ++m)
s[id1][m] = 0;
}
for(int m = k+1; m < j; ++m)//low
{
id2 = str[m] - 'a';
if(str[k] == '=' || str[k] == '+')
{
if(!s[id1][id2])
s[id1][26]++;
s[id1][id2] = 1;
}
else
{
if(s[id1][id2])
{
s[id1][id2] = 0;
s[id1][26]--;
}
}
}
}
i = j;
}
int num[26], same[27], m = 0;
for(i = 0; i < 26; ++i)
{
if(s[i][26] > 0)
num[m++] = i;
} // get exist
for(i = 0; i < m; ++i)
{
memset(same, 0, sizeof(same));
for(j = 0; j < 26; ++j)//record the low
{
if(s[num[i]][j] )
same[j] = 1;
}
int ok = 1;
for(j = i+1; j < m; ++j)//check if there is a same up
{
for(k = 0; k < 26; ++k)
{
if(s[num[j]][k])
{
if(!same[k])
{
ok = 0;
break;
}
}
else
{
if(same[k])
{
ok = 0;
break;
}
}
}
if(!ok)
break;
}
for(k = i; k < j; ++k)
printf("%c", 'A'+num[k]);
for(k = 0; k < 26; ++k)
if(same[k])
printf("%c", k + 'a');
i = j-1;
}
printf("\n");
}
return 0;
}