思路：n只有35， 将n份为2部分，一部分为前n/2个物品的取舍（取或不去）， 另一部分为剩下物品的取舍，复杂度为2^(n/2)，枚举左边的数，然后二分右边的数找到最优解，写lower_bound需要去重，手写二分就不需要了

AC代码：

#include "iostream"#include "iomanip"#include "string.h"#include "stack"#include "queue"#include "string"#include "vector"#include "set"#include "map"#include "algorithm"#include "stdio.h"#include "math.h"#pragma comment(linker, "/STACK:102400000,102400000")#define bug(x) cout<
     
      <<" "<<"UUUUU"<
      

vex1, vex2;int main(){ int n,m; scanf("%d%d",&n,&m); for(int i=1; i<=n; ++i){ scanf("%d",&a[i]); a[i]%=m; } int p=n>>1, q=n-p; for(int i=0; i<(1<