AGC020C Median Sum

原题链接：https://beta.atcoder.jp/contests/agc020/tasks/agc020_c

Median Sum

Problem Statement

You are given N N integers A1,A2,⋯,AN A 1 , A 2 , ⋯ , A N .

Consider the sums of all non-empty subsequences of A A . There are
2N−1 2 N − 1 such sums, an odd number.

Let the list of these sums in non-decreasing order be S1,S2,⋯,S2N−1 S 1 , S 2 , ⋯ , S 2 N − 1 .

Find the median of this list, S2N−1 S 2 N − 1 .

Constraints

1≤N≤2000 1 ≤ N ≤ 2000
1≤Ai≤2000 1 ≤ A i ≤ 2000
All input values are integers.

Input

Input is given from Standard Input in the following format:

N N

A1 A2 ⋯ AN A 1   A 2   ⋯   A N

Output

Print the median of the sorted list of the sums of all non-empty subsequences of
A A .

Sample Input 1

3
1 2 1

Sample Output 1

2

In this case, S=(1,1,2,2,3,3,4) S = ( 1 , 1 , 2 , 2 , 3 , 3 , 4 ) . Its median is S4=2 S 4 = 2 .

Sample Input 2

1
58

Sample Output 2

58

In this case, S=(58) S = ( 58 ) .

题目大意

输入一个数组 A1,A2,⋯,AN A 1 , A 2 , ⋯ , A N 。

考虑所有 2N−1 2 N − 1 个非空子集，每个子集的权值是包含的所有元素之和。问这 2N−1 2 N − 1 个非空子集的权值的中位数是什么？

题解

虽然题目不让我考虑空集，但我偏要考虑。。。

加上空集的话， 2N 2 N 个集合刚好可以两两匹配互为补集，这样 2N 2 N 个子集的中位数就是 sum2 s u m 2 ，去掉空集以后，中位数就是大于等于 sum2 s u m 2 的第一个子集。

题解

#include
using namespace std;
const int M=2005;
int n,x,sum;
bitsetdp;
void in(){scanf("%d",&n);}
void ac()
{
    dp[0]=1;
    for(int i=1;i<=n;++i)scanf("%d",&x),dp|=dp<for(int i=sum+1>>1;i<=sum;++i)if(dp[i])printf("%d",i),exit(0);
}
int main(){in();ac();}