154  Recycling
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154 : Recycling
Hi,
I'm getting WA on 154, can anyone give me some critical inputs/outputs? Is there any special case/tricks?
Thank you!
I'm getting WA on 154, can anyone give me some critical inputs/outputs? Is there any special case/tricks?
Thank you!
Thanks for replying.
..Yes i checked for both cases like 'e' and 'essence' etc.
My algo is like this:
Get input for city n, seperate the five bin's information, and store it as a data for city n. Get the data for all the cities in the same way. Also, count the number of the cities.
Now starting from the first city, check how many of it's five bin's data match with the other cities. If matches, increase counter of this city.
Finally, check which city has the highest counter and output its index+1 (city 1 is stored as element 0 of array of structs)
I've checked it by hand and seems to be working, but the WA keeps coming.
Thank you.
..Yes i checked for both cases like 'e' and 'essence' etc.
My algo is like this:
Get input for city n, seperate the five bin's information, and store it as a data for city n. Get the data for all the cities in the same way. Also, count the number of the cities.
Now starting from the first city, check how many of it's five bin's data match with the other cities. If matches, increase counter of this city.
Finally, check which city has the highest counter and output its index+1 (city 1 is stored as element 0 of array of structs)
I've checked it by hand and seems to be working, but the WA keeps coming.
Thank you.
WA 154
Can somebody give some input and output I'm getting WA and can't find my mistake.Somebody give some tests,please.
Thanks.
Thanks.
someone who like to solve informatic problems.
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650
Here is my program.
[pascal]label 1;
var r,o,y,g,b:array[0..2000] of char;
i,j,k,n,min,max,ss:longint;
s:string;
begin
repeat
i:=0;
repeat
i:=i+1;
readln(s);
if s[1]='e' then break;
if s[1]='#' then goto 1;
j:=1;
repeat
if s[j]='r' then r:=s[j+2];
if s[j]='o' then o:=s[j+2];
if s[j]='y' then y:=s[j+2];
if s[j]='g' then g:=s[j+2];
if s[j]='b' then b:=s[j+2];
j:=j+4;
until j>=19;
min:=maxlongint;
until s[1]='e';
n:=i1;
for i:=1 to n do
begin
max:=0;
ss:=0;
for j:=1 to n do
if i<>j then
begin
if ss>max then max:=ss;
ss:=0;
if r<>r[j] then ss:=ss+1;
if o<>o[j] then ss:=ss+1;
if y<>y[j] then ss:=ss+1;
if g<>g[j] then ss:=ss+1;
if b<>b[j] then ss:=ss+1;
end;
if min>max then
begin
min:=max;
k:=i;
end;
end;
writeln(k);
until s[1]='#';
1:
end.[/pascal]
[pascal]label 1;
var r,o,y,g,b:array[0..2000] of char;
i,j,k,n,min,max,ss:longint;
s:string;
begin
repeat
i:=0;
repeat
i:=i+1;
readln(s);
if s[1]='e' then break;
if s[1]='#' then goto 1;
j:=1;
repeat
if s[j]='r' then r:=s[j+2];
if s[j]='o' then o:=s[j+2];
if s[j]='y' then y:=s[j+2];
if s[j]='g' then g:=s[j+2];
if s[j]='b' then b:=s[j+2];
j:=j+4;
until j>=19;
min:=maxlongint;
until s[1]='e';
n:=i1;
for i:=1 to n do
begin
max:=0;
ss:=0;
for j:=1 to n do
if i<>j then
begin
if ss>max then max:=ss;
ss:=0;
if r<>r[j] then ss:=ss+1;
if o<>o[j] then ss:=ss+1;
if y<>y[j] then ss:=ss+1;
if g<>g[j] then ss:=ss+1;
if b<>b[j] then ss:=ss+1;
end;
if min>max then
begin
min:=max;
k:=i;
end;
end;
writeln(k);
until s[1]='#';
1:
end.[/pascal]
someone who like to solve informatic problems.
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650

 Learning poster
 Posts: 93
 Joined: Sun Jan 12, 2003 3:30 pm
I change my code and now it is giving right answer to your input but it is stil getting WA.
[pascal]label 1;
var r,o,y,g,b:array[0..2000] of char;
i,j,k,n,min,max,ss:longint;
s:string;
begin
repeat
i:=0;
repeat
i:=i+1;
readln(s);
if s[1]='e' then break;
if s[1]='#' then goto 1;
j:=1;
repeat
if s[j]='r' then r:=s[j+2];
if s[j]='o' then o:=s[j+2];
if s[j]='y' then y:=s[j+2];
if s[j]='g' then g:=s[j+2];
if s[j]='b' then b:=s[j+2];
j:=j+4;
until j>=19;
until s[1]='e';
n:=i1;
min:=maxlongint;
for i:=1 to n do
begin
max:=0;
ss:=0;
for j:=1 to n do
{ if i<>j then}
begin
ss:=0;
if r<>r[j] then ss:=ss+1;
if o<>o[j] then ss:=ss+1;
if y<>y[j] then ss:=ss+1;
if g<>g[j] then ss:=ss+1;
if b<>b[j] then ss:=ss+1;
{ writeln(i,' ',j,' ',ss);}
if ss>max then max:=ss;
end;
{ if ss>max then max:=ss;}
{ writeln(k,'k');}
{ writeln(i,' ',max);}
if min>max then
begin
min:=max;
k:=i;
end;
end;
writeln(k);
until s[1]='#';
1:
end.[/pascal]
[pascal]label 1;
var r,o,y,g,b:array[0..2000] of char;
i,j,k,n,min,max,ss:longint;
s:string;
begin
repeat
i:=0;
repeat
i:=i+1;
readln(s);
if s[1]='e' then break;
if s[1]='#' then goto 1;
j:=1;
repeat
if s[j]='r' then r:=s[j+2];
if s[j]='o' then o:=s[j+2];
if s[j]='y' then y:=s[j+2];
if s[j]='g' then g:=s[j+2];
if s[j]='b' then b:=s[j+2];
j:=j+4;
until j>=19;
until s[1]='e';
n:=i1;
min:=maxlongint;
for i:=1 to n do
begin
max:=0;
ss:=0;
for j:=1 to n do
{ if i<>j then}
begin
ss:=0;
if r<>r[j] then ss:=ss+1;
if o<>o[j] then ss:=ss+1;
if y<>y[j] then ss:=ss+1;
if g<>g[j] then ss:=ss+1;
if b<>b[j] then ss:=ss+1;
{ writeln(i,' ',j,' ',ss);}
if ss>max then max:=ss;
end;
{ if ss>max then max:=ss;}
{ writeln(k,'k');}
{ writeln(i,' ',max);}
if min>max then
begin
min:=max;
k:=i;
end;
end;
writeln(k);
until s[1]='#';
1:
end.[/pascal]
someone who like to solve informatic problems.
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650

 Learning poster
 Posts: 93
 Joined: Sun Jan 12, 2003 3:30 pm
Try this following input :
r/P,o/G,y/S,g/A,b/N
r/G,o/S,y/A,g/P,b/N
r/G,o/S,y/A,g/P,b/N
e
#
The correct one should produce :
2
while yours :
1
I guess you have missing the idea from the description :
"From these data he wishes to determine the city whose allocation scheme (if imposed on the rest of the country) would cause the least impact, that is would cause the smallest number of changes in the allocations of the other cities."
The idea is to find minimum total changes of allocation that should be made when a city is to be an example for others.
[pascal]
for i:=1 to n do
begin
max:=0;
ss:=0;
for j:=1 to n do
{ if i<>j then}
begin
ss:=0;
if r<>r[j] then ss:=ss+1;
if o<>o[j] then ss:=ss+1;
if y<>y[j] then ss:=ss+1;
if g<>g[j] then ss:=ss+1;
if b<>b[j] then ss:=ss+1;
{ writeln(i,' ',j,' ',ss);}
if ss>max then max:=ss;
end;
{ if ss>max then max:=ss;}
{ writeln(k,'k');}
{ writeln(i,' ',max);}
if min>max then
begin
min:=max;
k:=i;
end;
end;
[/pascal]
I saw from your code, that you didn't count that. Instead, for each city I, you search city J that have the largest number of difference from city I, so that the value of max and ss will never be larger then 5.
Consider the input above :
r/P,o/G,y/S,g/A,b/N
r/G,o/S,y/A,g/P,b/N
r/G,o/S,y/A,g/P,b/N
e
#
The number of total changes should be made if city 1 is to be winner is
4 + 4 = 8
THe number of total changes should be made if city 2 is to be winner is
4 + 0 = 4
The number of total changes should be made if city 3 is to be winner is
4 + 0 = 4,
thus the city with minimal number of changes is city 2
I hope it helps
r/P,o/G,y/S,g/A,b/N
r/G,o/S,y/A,g/P,b/N
r/G,o/S,y/A,g/P,b/N
e
#
The correct one should produce :
2
while yours :
1
I guess you have missing the idea from the description :
"From these data he wishes to determine the city whose allocation scheme (if imposed on the rest of the country) would cause the least impact, that is would cause the smallest number of changes in the allocations of the other cities."
The idea is to find minimum total changes of allocation that should be made when a city is to be an example for others.
[pascal]
for i:=1 to n do
begin
max:=0;
ss:=0;
for j:=1 to n do
{ if i<>j then}
begin
ss:=0;
if r<>r[j] then ss:=ss+1;
if o<>o[j] then ss:=ss+1;
if y<>y[j] then ss:=ss+1;
if g<>g[j] then ss:=ss+1;
if b<>b[j] then ss:=ss+1;
{ writeln(i,' ',j,' ',ss);}
if ss>max then max:=ss;
end;
{ if ss>max then max:=ss;}
{ writeln(k,'k');}
{ writeln(i,' ',max);}
if min>max then
begin
min:=max;
k:=i;
end;
end;
[/pascal]
I saw from your code, that you didn't count that. Instead, for each city I, you search city J that have the largest number of difference from city I, so that the value of max and ss will never be larger then 5.
Consider the input above :
r/P,o/G,y/S,g/A,b/N
r/G,o/S,y/A,g/P,b/N
r/G,o/S,y/A,g/P,b/N
e
#
The number of total changes should be made if city 1 is to be winner is
4 + 4 = 8
THe number of total changes should be made if city 2 is to be winner is
4 + 0 = 4
The number of total changes should be made if city 3 is to be winner is
4 + 0 = 4,
thus the city with minimal number of changes is city 2
I hope it helps
Finally I got it AC. You where right I had misunderstand the problem. Thank you for help.
Eduard
Eduard
someone who like to solve informatic problems.
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650
WA
Hello,
My program gives correct output for thses sample inputs, still i'm getting WA. Here is my code:
[c] Code Removed 07.08.2004[/c]
My program gives correct output for thses sample inputs, still i'm getting WA. Here is my code:
[c] Code Removed 07.08.2004[/c]

 Learning poster
 Posts: 93
 Joined: Sun Jan 12, 2003 3:30 pm
Thank you for replying,
But, shouldn't there be input for five wastes per line for each city. The problem statement says:
r/P,y/A,*/*,*/*,*/*
Please help.
Thank you,
But, shouldn't there be input for five wastes per line for each city. The problem statement says:
So, shouldn't there be 3 more c/w's in your input? LikeThe bins come in 5 different coloursred, orange, yellow, green and blueand 5 wastes have been identified for recyclingPlastic, Glass, Aluminium, Steel, and Newspaper.
r/P,y/A,*/*,*/*,*/*
Please help.
Thank you,

 Learning poster
 Posts: 93
 Joined: Sun Jan 12, 2003 3:30 pm
hehehe .. I'm sorry for my laziness, but actually my point is having :
r/P,o/G,y/S,g/A,b/N
is the same as having :
o/G,r/P,g/A,y/S,b/N
As you can see above, the important thing is the couple x/y, where x is color of the bin, and y is type of the waste,
thus, the position itself in the line doesn't matter
I hope it helps
r/P,o/G,y/S,g/A,b/N
is the same as having :
o/G,r/P,g/A,y/S,b/N
As you can see above, the important thing is the couple x/y, where x is color of the bin, and y is type of the waste,
thus, the position itself in the line doesn't matter
I hope it helps
154  Recycling
at first i put all information in a threedimensional array data[101][5][4]
so if the input is:
r/P,o/G,y/S,g/A,b/N
r/G,o/P,y/S,g/A,b/N
e
#
then
data[0][0] = "r/P",data[0][1] = "o/G",...
data[1][0] = "r/G",data[1][1] = "o/P",...
and i calculate the answer by compare every two countries,get the leastdifferent country
[cpp]
int calculate(char data[101][5][4],int n)
{
int min = 6,sum,record;
for(int i = 0;i < n;i++)
{
sum = 0;
for(int j = 0;j < n;j++)
{
if(j == i) continue;
for(int k = 0;k < 5;k++)
{
if(strcmp(data[k],data[j][k]) != 0)
{
sum++;
}
}
}
if(sum < min)
{
min = sum;
record = i;
}
}
return record + 1;
}
[/cpp]
so why would i get the WA???
so if the input is:
r/P,o/G,y/S,g/A,b/N
r/G,o/P,y/S,g/A,b/N
e
#
then
data[0][0] = "r/P",data[0][1] = "o/G",...
data[1][0] = "r/G",data[1][1] = "o/P",...
and i calculate the answer by compare every two countries,get the leastdifferent country
[cpp]
int calculate(char data[101][5][4],int n)
{
int min = 6,sum,record;
for(int i = 0;i < n;i++)
{
sum = 0;
for(int j = 0;j < n;j++)
{
if(j == i) continue;
for(int k = 0;k < 5;k++)
{
if(strcmp(data[k],data[j][k]) != 0)
{
sum++;
}
}
}
if(sum < min)
{
min = sum;
record = i;
}
}
return record + 1;
}
[/cpp]
so why would i get the WA???
"Learning without thought is useless；thought without learning is dangerous."
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"Hold what you really know and tell what you do not know this will lead to knowledge."Confucius