#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#define N 101

using namespace std;

int a[N],vis[N];

int n,m,ans1,ans2,mn,sum;

inline void check(int x,int y)

{

    if(mn>abs(x-y))

    {

        mn=abs(x-y);

        ans1=x,ans2=y;

    }

    return;

} 

void dfs(int now,int a1,int a2,int t1,int t2)

{

    if(now==n+ && abs(t1-t2)<= && a1+a2==sum) check(a1,a2);

    for(int i=now;i<=n;++i)

    {

        if(!vis[i])

        {

            vis[i]=;

            dfs(now+,a1+a[i],a2,t1+,t2);

            vis[i]=;

            dfs(now+,a1,a2+a[i],t1,t2+);

            vis[i]=;

        }

    }

    return;

}

int main()

{

    scanf("%d",&n);mn=0x3f3f3f3f;

    for(int i=;i<=n;++i) scanf("%d",&a[i]),sum+=a[i];

    dfs(,,,,);

    if(ans1>ans2) swap(ans1,ans2);

    printf("%d %d\n",ans1,ans2);

}

暴力dfs70

/*

二维费用背包模型

设dp[i][j][k]表示前i个人选j个能否得到体重为k。

这样dp的转移有2种，不选(dp[i-1][j][k])，选(dp[i-1][j-1][k-w[i]])(w表示体重),

得到dp数组以后，再枚举可能的答案，然后求的差值最小的，得到答案

*/

#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

#include<algorithm>

using namespace std;

bool dp[][];

int n,w[],s;

int main()

{

    scanf("%d",&n);

    for(int i = ;i <= n;i ++)  scanf("%d",&w[i]),s += w[i];

    dp[][] = ;

    int k = (n+)/;

    for(int i = ;i <= n;i ++)

      for(int j = k-;j >= ;j --)

        for(int v = *i;v >= ;v --)

            if(dp[j][v])

                dp[j+][v+w[i]] = ;

    int minn=<<;

    n = ;

    for(int i = ;i <= *k;i ++)

    {

        if(dp[k][i] && abs(s - i*) < minn) //看成s-i和i的差值

        {

            minn = abs(s-i*);

            if(i <= s/)    n = i;

            else    n = s-i;

        }

    }

    printf("%d %d",n,s-n);

    return ;

}