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11285 - Exchange Rates
Posted: Fri Dec 28, 2007 1:46 pm
by TimeString
i use O(n^2) solution but seems it does not expected, i got TLE.
it's very strange because the limit of n is 365, so i just want to check do i really have to implement an O(n) solution?
Posted: Fri Dec 28, 2007 2:42 pm
by rio
I can't imagine a O(n^2) solution. Could you post you code ?
-----
Rio
Posted: Fri Dec 28, 2007 2:50 pm
by TimeString
Code: Select all
#include <stdio.h>
int main(){
int en;
double erate[400];
int i,j;
double dp[400];
while(scanf("%d",&en)==1 && en>0){
for(i=1;i<=en;i++)
scanf("%lf",&erate[i]);
dp[0]=1000.0;
for(i=1;i<=en;i++){
dp[i]=0.0;
for(j=0;j<i;j++){
static double t;
t=dp[j]/erate[j]*0.97*erate[i]*0.97;
if(t>dp[i])
dp[i]=t;
if(dp[j]>dp[i])
dp[i]=dp[j];
}
}
printf("%.2lf\n",dp[en]);
}
return 0;
}
Posted: Fri Dec 28, 2007 3:41 pm
by rio
You code has a bug. You didn't initialize erate[0], but using in code:
Code: Select all
for(j=0;j<i;j++){
static double t;
t=dp[j]/erate[j]*0.97*erate[i]*0.97;
if(t>dp[i])
dp[i]=t;
if(dp[j]>dp[i])
dp[i]=dp[j];
}
I sent a code which initialize erate[0] with 1e9 and it got WA in 0.060 s.
Anyway, O(n) DP is way easier to implement.
-----
Rio
Posted: Fri Dec 28, 2007 4:04 pm
by TimeString
rio wrote:You code has a bug. You didn't initialize erate[0], but using in
I sent a code which initialize erate[0] with 1e9 and it got WA in 0.060 s.
thanks a lot.
but i think the bug will lead to runtime error or wrong answer, not time limited exceed = =||
Posted: Fri Dec 28, 2007 4:20 pm
by rio
Send a code with commenting out below part.
Code: Select all
if(t>dp[i])
dp[i]=t;
if(dp[j]>dp[i])
dp[i]=dp[j];
You could guess why its getting TLE.
-----
Rio
Posted: Sun Dec 30, 2007 8:14 am
by TimeString
rio wrote:Send a code with commenting out below part.
Code: Select all
if(t>dp[i])
dp[i]=t;
if(dp[j]>dp[i])
dp[i]=dp[j];
You could guess why its getting TLE.
i still don't know why.
does it mean that it costs a lot of time to do float number comparison?