本题不难,但是由于没有好的算术方案,所以编的时间长,又因为pascal递归不稳定,所以调试时间长,提交了三次.
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 204 KB]
Test 2: TEST OK [0.000 secs, 204 KB]
Test 3: TEST OK [0.000 secs, 208 KB]
Test 4: TEST OK [0.000 secs, 208 KB]
Test 5: TEST OK [0.000 secs, 204 KB]
Test 6: TEST OK [0.011 secs, 208 KB]
Test 7: TEST OK [0.011 secs, 208 KB]
All tests OK.
Your program ('zerosum') produced all
correct answers! This is
your
submission #3 for this
problem.
Congratulations!
program
test;
type
node=array[0..8]of integer;
var
n:3..9;
a:array[1..9]of integer;
b1:array[1..9]of integer;
total:longint;
b:node;
procedure
print(c:node);
var
i:integer;
begin
for i:=2 to n do
begin
case c[i-1] of
1:begin
write(a[i-1],' ');
end;
2:begin
write(a[i-1],'+');
end;
3:begin
write(a[i-1],'-');
end;
end;
end;
writeln(a[n]);
{for i:=1 to n-1 do
write(c[i],' ');
writeln;}
end;
procedure
pd(c:node);
var
i,temp,temp2:integer;
begin
fillchar(b1,sizeof(b1),0);
total:=0;temp:=0;
for i:=1 to n do
begin
if c[i]=1 then
begin
if c[i-1]=1 then total:=total*10+a[i+1];
if c[i-1]<>1 then
begin
total:=a[i];
total:=total*10+a[i+1];
end;
if c[i+1]<>1 then
begin inc(temp);b1[temp]:=total;end;
end;
end;
temp:=1;temp2:=0;
if c[1]=1 then begin total:=b1[1];inc(temp2);end
else total:=a[1];
while temp<n do
begin
if c[temp]<>1 then
begin
if c[temp+1]<>1 then
case c[temp] of
2:total:=total+a[temp+1];
3:total:=total-a[temp+1];
end
else
begin
inc(temp2);
case c[temp] of
2:total:=total+b1[temp2];
3:total:=total-b1[temp2];
end;
for i:=temp+1 to n do
if i<>1 then
begin
temp:=i;
break;
end;
end;
end;
inc(temp);
end;
if total=0 then print(c)
else exit;
end;
procedure
dfs(x:integer;b:node);
var
i:integer;
begin
if (x=n) then
begin
pd(b);
exit;
end;
for i:=1 to 3 do
begin
b[x]:=i;
if x+1<=n then dfs(x+1,b);
end;
end;
procedure
init;
var
i:integer;
begin
readln(n);
fillchar(a,sizeof(a),0);
for i:=1 to n do
a[i]:=i;
b[0]:=0;
end;
begin
assign(input,'zerosum.in');
reset(input);
assign(output,'zerosum.out');
rewrite(output);
init;
dfs(1,b);
close(input);
close(output);
end.
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