Dividing coins 动规
题目：
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
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 (m 100) 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
题目大意：
给n个数，求这些数分成两拨，求这两拨差值最小的值
题目思路：
1、因为n最大为100，如果单纯用搜索会超时，所以可以用动态规划来解决
2、dp【求第几个硬币i】【差值j】=dp【i-1】【j+fei【i】】||dp【i】【j】
3.
程序：
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cctype>
#include <fstream>
#include <limits>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cassert>
using namespace std;
int a[200];
int main()
{
int ci;
scanf("%d",&ci);
while(ci--)
{
memset(a,0,sizeof(a));
int sum=0,n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
set<int>se[2];
set<int>::iterator it;
se[0].insert(0);
for(int i=0;i<n;i++)
{
int f=a[i];
for(it=se[i%2].begin();it!=se[i%2].end();it++)
{
int t=*it;
se[(i+1)%2].insert(t+f);
se[(i+1)%2].insert(abs(t-f));
}
se[i%2].clear();
}
it=se[n%2].begin();
cout<<*it<<endl;
}
return 0;
}