CF1703G Good Key, Bad Key

Good Key, Bad Key - 洛谷

G. Good Key, Bad Key

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are nn chests. The ii-th chest contains aiai coins. You need to open all nn chests in order from chest 11 to chest nn.

There are two types of keys you can use to open a chest:

• a good key, which costs kk coins to use;
• a bad key, which does not cost any coins, but will halve all the coins in each unopened chest, including the chest it is about to open. The halving operation will round down to the nearest integer for each chest halved. In other words using a bad key to open chest ii will do ai=⌊ai2⌋ai=⌊ai2⌋, ai+1=⌊ai+12⌋,…,an=⌊an2⌋ai+1=⌊ai+12⌋,…,an=⌊an2⌋;
• any key (both good and bad) breaks after a usage, that is, it is a one-time use.

You need to use in total nn keys, one for each chest. Initially, you have no coins and no keys. If you want to use a good key, then you need to buy it.

During the process, you are allowed to go into debt; for example, if you have 11 coin, you are allowed to buy a good key worth k=3k=3 coins, and your balance will become −2−2 coins.

Find the maximum number of coins you can have after opening all nn chests in order from chest 11 to chest nn.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The first line of each test case contains two integers nn and kk (1≤n≤1051≤n≤105; 0≤k≤1090≤k≤109) — the number of chests and the cost of a good key respectively.

The second line of each test case contains nn integers aiai (0≤ai≤1090≤ai≤109)  — the amount of coins in each chest.

The sum of nn over all test cases does not exceed 105105.

Output

For each test case output a single integer  — the maximum number of coins you can obtain after opening the chests in order from chest 11 to chest nn.

Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).

Example

input

Copy

5
4 5
10 10 3 1
1 2
1
3 12
10 10 29
12 51
5 74 89 45 18 69 67 67 11 96 23 59
2 57
85 60

output

Copy

11
0
13
60
58

Note

In the first test case, one possible strategy is as follows:

• Buy a good key for 55 coins, and open chest 11, receiving 1010 coins. Your current balance is 0+10−5=50+10−5=5 coins.
• Buy a good key for 55 coins, and open chest 22, receiving 1010 coins. Your current balance is 5+10−5=105+10−5=10 coins.
• Use a bad key and open chest 33. As a result of using a bad key, the number of coins in chest 33 becomes ⌊32⌋=1⌊32⌋=1, and the number of coins in chest 44 becomes ⌊12⌋=0⌊12⌋=0. Your current balance is 10+1=1110+1=11.
• Use a bad key and open chest 44. As a result of using a bad key, the number of coins in chest 44 becomes ⌊02⌋=0⌊02⌋=0. Your current balance is 11+0=1111+0=11.

At the end of the process, you have 1111 coins, which can be proven to be maximal.

=========================================================================

DP问题，常规的想即可，唯一需要注意的是，当坏钥匙用超过30次的时候，虽然再次使用坏钥匙无法降低物品价值，但是却可以比用好钥匙更优，这时候一定不能用好钥匙，否则就是花费k获得0的价值，dp[i][x]=......=..dp[i][32]=dp[i][31]=dp[i][30]统一把x设成35即可，代表超过30次坏钥匙的情况。

# include
# include
 
using namespace std;
typedef long long int ll;
 
ll dp[100000+10][50];
 
ll b[100000+10];
 
ll a[100000+10][50];
 
int main ()
{
 
    int t;
 
    cin>>t;
 
    while(t--)
    {
        ll n,k;
 
 
        scanf("%lld%lld",&n,&k);
 
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&b[i]);
 
        }
 
        for(int i=1; i<=n; i++)
        {
            a[i][0]=b[i];
 
            for(int j=1; j<=35; j++)
            {
                a[i][j]=a[i][j-1]/2;
 
            }
        }
 
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<=35; j++)
            {
                dp[i][j]=-1e18;
 
            }
        }
 
        ll ans=-1e18;
 
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<=35; j++)
            {
                if(j>=1)
                dp[i][j]=max(dp[i-1][j]+a[i][j]-k,dp[i-1][j-1]+a[i][j]);
                else
                dp[i][j]=dp[i-1][j]+a[i][j]-k;
                
                if(j==35)
                    dp[i][j]=max(dp[i][j],dp[i-1][35]);
                
                
 
                if(i==n)
              ans=max(ans,dp[n][j]);
                
            }
 
 
        }
 
        cout<
    }
 
 
    return 0;
 
 
}