gym 101061F. Fairness(dfs或者dp）

题目链接：http://codeforces.com/gym/101061/problem/F

题面：

F. Fairness
time limit per test2.0 s
memory limit per test64 MB
inputstandard input
outputstandard output
Dwik and his brother Samir both received scholarships from a famous university in India. Their father, Besher, wants to send some money with each of them.

Besher has n coins, the ith coin has a value of ai. He will distribute these coins between his two sons in n steps. In the ith step, he chooses whether to give the ith coin to Dwik or to Samir.

Let xi be the absolute difference between the sum of Dwik’s and Samir’s coins after the ith step. The unfairness factor of a distribution is max({x1, x2, …, xn}). Besher wants to minimize the unfairness factor, can you help him?

Input
The first line of the input consists of a single integer t, the number of test cases. Each test case consists of 2 lines:

The first line contains an integer n (1 ≤ n ≤ 100).

The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 100).

Output
Print t lines, ith line containing a single integer, the answer to the ith test case.

Example
inputCopy
2
5
1 2 1 4 3
7
4 5 6 1 1 3 4
outputCopy
2
5
Note
In the first sample test, besher has 5 coins (1, 2, 1, 4, 3), he can distribute them in the following way:

Step 1: Give the first coin to dwik , d = 1, s = 0 x1 = |1 - 0| = 1

Step 2: Give the second coin to samir, d = 1, s = 2 x2 = |1 - 2| = 1

Step 3: Give the third coin to samir, d = 1, s = 3 x3 = |1 - 3| = 2

Step 4: Give the fourth coin to dwik, d = 5, s = 3 x4 = |5 - 3| = 2

Step 5: Give the fifth coin to samir, d = 5, s = 6 x5 = |5 - 6| = 1

max({x1, x2, x3, x4, x5}) = 2

---

递归写法

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;
#define ll long long 
const int N=1e5+100;
const int mod=1000000007;

int a[110];

int sum[110];
int mx;
int n;

void dfs(int pos,int sm,int cnt)
{
	if(cnt>=mx) return ;
	if(pos>=n)
	{
		if(cnt<mx) mx=cnt;
		cnt=0;
		return ;
	}
	int tem=abs(sum[pos]-sm-a[pos] -sm-a[pos] );
	if(tem<mx)
	{
		dfs(pos+1,sm+a[pos],cnt>=tem?cnt:tem);
	}
	if(pos!=0)
	{
		tem=abs(2*sm -sum[pos]);
		if(tem<mx)
		{
			dfs(pos+1,sm,cnt>=tem?cnt:tem);
		}
	}
	return ;
	
		
}
int main()
{
	// freopen("data.in","r",stdin);
	// ios::sync_with_stdio(false);
	// cin.tie(0);
	// cout.tie(0);
	int t;
	//cin>>t;
	scanf("%d",&t);
	
	while(t--)
	{

		//cin>>n;
		scanf("%d",&n);
		mx=0;
		for(int i=0;i<n;i++)
		{
			cin>>a[i];
			if(a[i]>mx) mx=a[i];
			if(i==0) 
				sum[i]=a[i];
			else
				sum[i]=sum[i-1]+a[i];
		}
		dfs(0,0,0);
		//cout<<mx<<endl;
		printf("%d\n",mx);
		
	}
	
	return 0;
}

dp写法