## 10014 - Simple calculations

All about problems in Volume 100. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

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stcheung
Experienced poster
Posts: 114
Joined: Mon Nov 18, 2002 6:48 am
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### Re: 10014 - Simple Calculations

aeiou, consider the following input. Your program returns 34.62 while the answer should be 27.85. By the way, if you want an alternative way to do it, you can consider writing each a in terms of a[1], a[0] and a constant. For instance, you can write a[2] as 2*(a[1]+c[1]) - a[0] and so forth until you can formulate a[n+1] in terms of a[1], a[0] and a constant. Then just solve for a[1].

1

2
50.50
43.45
10.15
10.15

raihan004
New poster
Posts: 2
Joined: Tue Oct 09, 2012 2:07 am

### 10014 got time limit error

here is my code face time limt error how solve this problem
using this logic?

Code: Select all

``````#include<stdio.h>
int rec(long n)
{
if(n%10>0)
return n%10;
else if(n==0)
return 0;
else rec(n/10);
}
int main()
{
long p,q;
long sum=0;
long i;
while(scanf("%ld%ld",&p,&q)!=0)
{
if(p<0&&q<0)
break;
else
{
for(i=p;i<=q;i++)
sum+=rec(i);
}
printf("%ld\n",sum);
sum=0;
}
return 0;
}

``````

brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA

### Re: 10014 got time limit error

It looks like the wrong problem number. Try to think of a faster method. For example, if you were asked to sum the numbers from 1 to 1000000 would you iterate through them all?
Check input and AC output for thousands of problems on uDebug!

shondhi
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Posts: 25
Joined: Tue Oct 02, 2012 5:24 pm
Location: Chittagong
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### 10014 - Simple calculations

I can't understand why I got WA? Can anyone please help me. Here is my code:

Code: Select all

``````#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n, test, i;
double sum;
double a1, a0, a2;
double ci[3002];
while(scanf("%d", &n) == 1)
{
cout<<endl;
sum = 0;
scanf("%d", &n);
scanf("%lf", &a0);
scanf("%lf", &a2);
for(i = 1; i <= n; i++)
{
scanf("%lf", &ci[i]);
}
for(i = 1; i <= n; i++)
{
sum += ci[i];
}
a1 = ((a0 + a2)/2) - sum;
printf("%0.2lf\n", a1);
}
return 0;
}
``````
N.B: Code will be removed after AC.

brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA

### Re: 10014 - Simple calculations

Check input and AC output for thousands of problems on uDebug!