hdu6092 01背包

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 658    Accepted Submission(s): 297

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has n

positive A1−An

and their sum is m

. Then for each subset S

of A

, Yuta calculates the sum of S

.

Now, Yuta has got 2n

numbers between [0,m]

. For each i∈[0,m]

, he counts the number of i

s he got as Bi

.

Yuta shows Rikka the array Bi

and he wants Rikka to restore A1−An

.

It is too difficult for Rikka. Can you help her?

 

Input

The first line contains a number t(1≤t≤70)

, the number of the testcases.

For each testcase, the first line contains two numbers n,m(1≤n≤50,1≤m≤104)

.

The second line contains m+1

numbers B0−Bm(0≤Bi≤2n)

.

 

Output

For each testcase, print a single line with n

numbers A1−An

.

It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.

 

Sample Input

2

2 3

1 1 1 1

3 3

1 3 3 1

 

Sample Output

1 2

1 1 1

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e4+;
typedef long long LL;
LL dp[N],B[N];
int a[N];
int main(){
    int n,m,T;
    for(scanf("%d",&T);T--;){
    memset(dp,,sizeof(dp));
    memset(a,,sizeof(a));
    scanf("%d%d",&n,&m);
    for(int i=;i<=m;++i) scanf("%I64d",&B[i]);
    dp[]=;
    for(int i=;i<=m;++i){
        if(dp[i]==B[i]) continue;
        a[i]=B[i]-dp[i];
        for(int j=;j<=a[i];++j) for(int k=m;k>=i;--k) dp[k]+=dp[k-i];
    }
    int i;
    for(i=;i<=m;++i) if(a[i]--) {printf("%d",i);break;}
    for(;i<=m;++i) while(a[i]--) printf(" %d",i);
    puts("");
    }
}