Description
X is fighting beasts in the forest, in order to have a better chance to survive he's gonna buy upgrades for his weapon. Weapon upgrade shops are available along the forest, there are n shops, where the ith of them provides an upgrade with energy ai. Unfortunately X can make use only of the maximum power of two that divides the upgrade energy, lets call it the powerincrement. For example, an upgrade with energy of 6 has a power increment of 1 because 6 is divisible by 21 but not divisible by 22 or more, while for upgrade with energy of 5 power increment is 0, also after buying a weapon with energy v, X can only buy upgrades with energies greater than v, in other words, the upgrades energies that X is gonna buy should be in strictly increasing order. X is wondering what is the maximum power he can achieve at the end of his journey. note that only the energy of upgrades should be in strictly increasing order not the power increment of the upgrade. 1 < n < 100, 1 ≤ ai ≤ 106
The first line of input contains one integer, T, which denotes number of test cases. Each test case contains two lines, the first one contains one integer n which denotes number of shops, and the next line contains n integers the ith of them denotes the energy of the upgrade provided by the ith shop.
Print one integer denotes maximum power X can achieve at the end of his journey
2
3
1 10 16
2
8 12
5
5
题意：商店有ai的能量，主角升级按照ai能被2^n的整除（n要最大）（比如8可以被8整除（2^3），则升级3级），而且获得的能量是递增的（也就是如果 8 10 7 12，如果选了8，以后7就不能选了）
最多能升级多少
解法：一开始我们就固定最大值，比如固定8为最大值，算出加到8可以升级多少（这里它是第一个则计算自己），然后固定10，最后固定12
dp[i]=max(dp[i],dp[j]+b[i])(i是计算到自己时一共能升级多少，j<i且相加按照a[i] > a[j]的顺序)
#include<bits/stdc++.h>
using namespace std;
int a[10000],b[10000];
int dp[10000];
int main()
{
int t;
cin>>t;
while(t--)
{
memset(dp,0,sizeof(dp));
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
int num=0;
int sum=1;
cin>>a[i];
while(a[i]%sum==0)
{
num++;
sum*=2;
}
b[i]=num-1;
dp[i]=num-1;
// cout<<num-1<<endl;
}
for(int i = 1;i <= n;i++)
{
for(int j = 0;j < i;j++)
{
if(a[i] > a[j])
{
dp[i] = max(dp[i],b[i] + dp[j]);
}
}
}
int ans=0;
for(int i=1; i<=n; i++)
ans=max(ans,dp[i]);
printf("%d\n",ans);
}
return 0;
}