OJ Board

oh.. im sorry.. here's the input i intended to post correct answer would be 4. but your code gives 3.

naseef, check your output for following input.

hmm.. i reduced my matrix multiplication part from 2*n^3 to n^3.. and now getting WA
im doing all calculation on unsigned long without any mod operation and its giving me right answer in my compiler.. is it wrong ? should i use mod operation ?

im using matrix expo but it seems to be slow on input sets

algo:
expo( p )
{
if( p==1 )
return base
else
{
a = expo(p/2);
b = a*a;
if(p%2==1)
a = a*base
return a;
}

so, for odd values of "p" im doing 2 matrix multiplication. so my algorithms complexity is actually 2*3^n*log(p). any ...

after reading taman's post, i looked at my code again and cant decide whether it would be dp or greedy. but i think its a close resembelance to LIS algorithm with a bit of greedy choice. sorry if it makes anyone misguided. :(
here's my algo,

1. sort (criteria is yours to decide)
2. now, for i'th ...

greedy will do..

can anyone give me any hint.

@tanmoy, i think all of the graph in your i/o is not DAG. but strangly, in problem statement, 2nd graph contains a self loop. although i treated given graph as a DAG and got AC.

there's an easy problem which is quite similar with this one, 562 : dividing coins.

for PI use this and for floating number comparison use

its a DP problem...

sometime wrote:Have your 10123 - No Tipping Accepted? My code is Time Limit Exceeded.
Could you help me? Thank you.
anyone help me?? thank you very much~~~

i used memoization to solve this task. think abt it. if u hav trouble then PM me

i am unsuccessfully trying to solve this problem.... here's my algo

1. generate digit[x] -> number of digit(s) in x^x ( using log )
2. take input and binary search in digit[] to get a approx
3. use fast exponentiation to evaluate x^x
4. check whether calculated x^x is equal as given input string ...

for input 10000 100, ur output can be different than ACed output posted by red scorpion. i solved this problem using 3 different equation and all of them produced different output for above case but all of those code have gotten AC.