2008 Canadian Computing Competition, Day 2

Day 2, Problem 2: Candy

You and a friend have a big bag of candy. You want to keep slim and trim, and so you would like to equalize the candy which you are sharing with your friend in terms of calorie count. That is, your task is to divide the candies into two groups such that the number of calories in each group is as close together as possible.

Input

The first line of input contains the number of different kinds of candy you have in your bag of candy N (1 ≤ N ≤ 100). On the following N lines, there are pairs of numbers describing each type of candy. The candy description is of the form k_i c_i where k_i is the number of that particular type of candy contained in the bag and c_i is the calorie count for each piece of that type of candy. You may assume that 1 ≤ k_i ≤ 500 and 1 ≤ c_i ≤ 200.

Output

Your output is one integer which is the minimum difference of calories between friends

Sample Input

4
3 5
3 3
1 2
3 100

Sample Output

74

Explanation

Your friend takes two of the 100-calorie candies, for a total of 200 calories. You keep the remaining candies, which have 126 calories.

Grading

You may assume that 50% of the test cases will have at 1 ≤ N, k_i, c_i ≤ 100. All test cases will have 1 ≤ N ≤ 100, 1 ≤ k_i ≤ 500 and 1 ≤ c_i ≤ 200. You solution must use at most 512MB of memory and run in at most 6 seconds.

All Submissions
Best Solutions

Point Value: 20
Time Limit: 6.00s
Memory Limit: 512M
Added: Dec 07, 2008

Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3

Comments (Search)

i'm assuming accessing memory is much faster than calculating expressions in for loops?

Are you talking about something like replacing
[code]
for (i=0; i<x+y; i++)
...
[/code]
with
[code]
int z=x+y;
for (i=0; i<z; i++)
...
[/code]
?
Obviously the first is slower than the second, but usually this isn't the difference between AC and TLE...

 
for (i=0; i<x+y; i++) 
... 

 
int z=x+y; 
for (i=0; i<z; i++) 
...