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?
11285 - Exchange Rates
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- Joined: Mon Nov 13, 2006 3:53 am
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;
}
You code has a bug. You didn't initialize erate[0], but using in code:
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.
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Rio
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];
}
Anyway, O(n) DP is way easier to implement.
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Rio
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Send a code with commenting out below part.
You could guess why its getting TLE.
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Rio
Code: Select all
if(t>dp[i])
dp[i]=t;
if(dp[j]>dp[i])
dp[i]=dp[j];
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Rio
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- New poster
- Posts: 26
- Joined: Mon Nov 13, 2006 3:53 am
i still don't know why.rio wrote:Send a code with commenting out below part.You could guess why its getting TLE.Code: Select all
if(t>dp[i]) dp[i]=t; if(dp[j]>dp[i]) dp[i]=dp[j];
does it mean that it costs a lot of time to do float number comparison?