371 - Ackermann Functions

[Aengus]

It's funny, that everything is exactly the same that I've thought. May be somebody can check the code:
[c]#include <stdio.h>

unsigned next(unsigned a) {
if (a == 1) return 0;

if (a % 2 == 0) return a / 2;
else return 3*a + 1;
}

int seq_len(unsigned a) {
int steps = 0;

if (a == 1) return 3;

while (a = next(a)) steps++;

return steps;
}

int main(void) {
unsigned l, h, t, i, max, v, len;

while (1) {
scanf("%u %u", &l, &h);
if (l == 0) break;

printf("Between %lu and %lu, ", l, h);

if (l > h) {
t = l; l = h; h = t;
}

max = 0; v = 0;
for (i = l; i <= h; i++) {
len = seq_len(i);
if (len > max) {
max = len;
v = i;
}
}

printf("%lu generates the longest sequence of %d values.\n", v, max);
}

return 0;
}[/c]
[/c]

[osan]

Dear Aengus

Check these

1 2
Between 1 and 2, 1 generates the longest sequence of 3 values.
2 1
Between 1 and 2, 1 generates the longest sequence of 3 values.

[Aengus]

Osan,

Thank you for your answer, but I tried not to swap values if first is greater than the second and nothing changed. Besides, it is not clear from the statement of the problem, what to output in case of L>H. The sequence starting with 1 has length 3 (it is the example in the problem), the sequence starting with 2 has length 2, so it should be right.

I want to add, that I've spent quite a lot of time on that problem. First I tried the code that I showed in my last message. Than I added a check that the next value will not be greater than 2^32-1 similar to one in the problem 694 (which I succesfuly solved). I've got runtime error that ment that the sequence members may grow beyond the limit of unsigned long. Then I've used several functions for long arithmetic (which I also succesfuly used in several other problems). Now I could get the answer for any interval within 1 .. 2^32-1. But I once again got WA for that solution. I can't figure out any problems with any of my solutions.

[osan]

Aengus,

I think you didn

[osan]

onece again telling you remove your code from the forum.

[osan]

Aengus,

I think you didn

[Aengus]

Oh, thank you! The trouble was really in that point.

[seadoo]

Hello all!

Why am I getting SIGSEV error with this piece of code?
The largest value in the sequence will not be larger than a 32-bit C long, and I have long's in the code. What else could be wrong?

[seadoo]

Got it... changed to unsigned long and it worked...

[mohiul alam prince]

this is my code why i am getting TLE
BUT i have solved 100 & 694
[cpp]
#include <stdio.h>
#include <math.h>

double cycle(long m)
{
double i = 0;

if(m==1)return 3;
while (m != 1){
if (m % 2 == 0){
m = m/2;
}
else{
m = 3*m+1;
}
i++;
}
return i;
}


int main()
{
long m,n,x,x1;
long mOriginal,nOriginal;
double temp,max,max1,min;
long i;

while (scanf("%ld %ld",&m,&n)==2){

if(n==0&&m==0)break;
mOriginal = m;
nOriginal = n;

if (m > n){
temp = m;
m = n;
n = temp;
}

max = cycle(m);
x=m;
for(i=m+1;i<=n;i++) {
temp = cycle(i);
if (temp > max){
max = temp;
x=m;
}
}
printf("Between %ld and %ld, %ld generates the longest sequence of %.0lf values.\n",mOriginal,nOriginal,x,max);

}
return 0;
}[/cpp][/cpp][/c]

[osan]

long's range 2^31-1
in the problem they told about 32 bit number.
how can u hold the number in long???
u can use unsigned long
or double

[mohiul alam prince]

i have used double don't u see it?
any thing ????????

[sohel]

Why use double when you can use long long.
I used long long and got AC.

you have used double in the wrong place.
The cycle length will not be that huge but
the value 3n + 1 might be.

Make the appropriate modification and get AC.

[mohiul alam prince]

the range of long long & double are equal
and i have used long long for all variables but still TLE

[mohiul alam prince]

sohel
whats ur algo can u explain
i have solve 694 & 100
but i am getting TLE