题目链接：https://codeforces.com/problemset/problem/1437/C

用时少的一定比用时大的先拿更优，所以按用时大小排个序，
然后按顺序 \(dp\),用时最多不会超过 \(2*n\)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 210;

int T, n;
int a[maxn], f[maxn][maxn*2];

struct P{
	int t,id;
}p[maxn];

bool cmp(P x, P y){ return x.t < y.t; }

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	T = read();
	while(T--){
		memset(f,0x3f,sizeof(f));
		int ans = f[0][0];
		n = read();
		for(int i=1;i<=n;++i) p[i].t = read(), p[i].id = i;
		sort(p+1,p+1+n,cmp);

		for(int i=1;i<=n;++i) printf("%d ",p[i].t); printf("\n");

		for(int j=1;j<=2*n;++j) f[1][j] = abs(p[1].t - j);
		for(int i=1;i<=n;++i){
			for(int j=1;j<2*n;++j){
				f[i][j+1] = min(f[i][j+1],f[i-1][j] + (j+1-p[i].t));
			}
		}
		
		for(int i=1;i<=n;++i){
			printf("%d:",i);
			for(int j=1;j<=2*n;++j){
				printf("%d ",f[i][j]);
			}printf("\n");
		}
		
		for(int j=p[n].t;j<=2*n;++j){
			ans = min(ans, f[n][j]); 
		} 
		printf("%d\n",ans);
	}
	return 0;
}