It's commonly known that the Dutch have invented copper-wire. Two Dutch menwere fighting over a nickel, which was made of copper. They were both soeager to get it and the fighting was so fierce, they stretched the coin togreat 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 verylast cent. Nowadays fighting over a single cent will not be seen anymore, butbeing capable of making an equal division as fair as possible is somethingthat will remain important forever...
That's what this whole problem is about. Not everyone is capable of seeinginstantly 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 divisionbetween two persons. This means that the difference between the amount eachperson obtains should be minimised. The value of a coin varies from 1 centto 500 cents. It's not allowed to split
a single coin.
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.
The output consists of n lines. Each line contains the minimal positivedifference between the amount the two persons obtain when they divide thecoins from the corresponding bag.
2 3 2 3 5 4 1 2 4 6
0 1
题目意思是有n个硬币，每个硬币有他的价值，分给2个人，使得2个人分得的硬币的价值尽量相等，输出最小差值。
解题思路：先求硬币的总价值，然后转化分背包问题，求用所有的硬币能取得的<=s/2的最大价值m。他们之间的差值最小即为s-2*m;
#include<stdio.h>
#include<string.h>
#define MAXN 101
int c[MAXN];
int f[50000];
int main(void)
{
int T;
// freopen("d:\\in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
int m,j;
scanf("%d",&m);
int i,s=0,tmp;
for(i=1;i<=m;i++)
{
scanf("%d",&c[i]);
s+=c[i];
}
tmp=s/2;
memset(f,0,sizeof(f));
f[0]=1;
for(i=1;i<=m;i++)
for(j=tmp;j>=c[i];j--)
{
if(f[j-c[i]])
f[j]=1;
}
j=tmp;
while(!f[j])
j--;
printf("%d\n",s-2*j);
}
return 0;
}