AYITACM2016省赛第二周 I 分硬币（01背包）

Description

It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great length and thus created copper-wire.

Not commonly known is that the fighting started, after the two Dutch tried to divide a bag with coins between the two of them. The contents of the bag appeared not to be equally divisible. The Dutch of the past couldn't stand the fact that a division should favour one of them and they always wanted a fair share to the very last cent. Nowadays fighting over a single cent will not be seen anymore, but being capable of making an equal division as fair as possible is something that will remain important forever...

That's what this whole problem is about. Not everyone is capable of seeing instantly what's the most fair division of a bag of coins between two persons. Your help is asked to solve this problem.

Given a bag with a maximum of 100 coins, determine the most fair division between two persons. This means that the difference between the amount each person obtains should be minimised. The value of a coin varies from 1 cent to 500 cents. It's not allowed to split a single coin.

Input 

A line with the number of problems  n , followed by  n  times:

• a line with a non negative integer m () indicating the number of coins in the bag
• a line with m numbers separated by one space, each number indicates the value of a coin.

Output 

The output consists of  n  lines. Each line contains the minimal positive difference between the amount the two persons obtain when they divide the coins from the corresponding bag.

Sample Input 

2
3
2 3 5
4
1 2 4 6

Sample Output 

0
1

分析：

首先输入n，代表有n组测试数据。每组数据2行，第一行为正整数m(0<=m<=100)，代表有m个硬币，第二行m个正整数ci(1<=ci<=500)，代表每个硬币的面值。求把硬币分给2个人，使得他们所得的总金额差值最小。

01背包判断可行性。定义状态d(i,j)，代表用前i种硬币组合出金额j的可行性，值为1代表可行，为0代表不行。

最后遍历所有可行状态，找出最小差值。

#include
#include
int f[1100000];
int main()
{
    int n,m,a[110],i,j,t,s;
    scanf("%d",&n);
    while(n--)
    {
        s=0;
        memset(f,0,sizeof(f));
        scanf("%d",&m);
        for(i=0; i=a[i]; j--) //只要大于这个硬币的包才能放这个硬币
                if(f[j-a[i]])  //表明用前i个硬币可以装满这个包
                    f[j]=1;
        for(i=s/2; i>=0; i--)//注意，可能只有一个硬币
            if(f[i]==1)
            {
                t=i;  //此时，另一方得到的钱数为0
                break; //钱数差就为s
            }
        printf("%d\n",s-2*t);
    }
    return 0;
}