2014
02-27

# Treasure Division

Ant Tony and Monkey Luffy have discovered a big treasure.It turns out to be a box of gold coins.There are N gold coins in the box.However,the gold coins are not all the same.Every gold coin has a value Vi.Now they wants to divide this big treasure into two halves.The number of gold coin differ in two half must be no more than one.Tony and Luffy wants to know the fairest division,ie,the division which minimize the difference between each parts.

Input contains multiple cases.Test cases are separated by several blank lines.

Each test case starts with a integer N(1<=N<=30) ,indicating that there are N gold coins int the treasure box.Followed by one line contain N integer,the ith integer vi (1<=vi<=2^30),representing the value of coin i.

Input contains multiple cases.Test cases are separated by several blank lines.

Each test case starts with a integer N(1<=N<=30) ,indicating that there are N gold coins int the treasure box.Followed by one line contain N integer,the ith integer vi (1<=vi<=2^30),representing the value of coin i.

3
2 2 4
4
1 2 3 6

0
2

Hint

In sample 1,we can divide the 3 gold coins into {1,2},{3},
their difference is 0.
In sample 2,we can divide the 4 gold coins into {1,4},{2,3},
their difference is 2,which is also the least difference that could be made.



http://acm.hit.edu.cn/hoj/problem/view?id=3017

#include <stdio.h>
#include <string.h>

int main()
{

{
{
printf("hit\n");
}
else
{
printf("lose\n");
}
}

return 0;
}

1. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;