2016 Al-Baath University Training Camp Contest-1 J

Description

Xisfightingbeastsintheforest,inordertohaveabetterchancetosurvivehe'sgonnabuyupgradesforhisweapon.Weaponupgradeshopsareavailablealongtheforest,therearenshops,wheretheithofthemprovidesanupgradewithenergyai.UnfortunatelyXcanmakeuseonlyofthemaximumpoweroftwothatdividestheupgradeenergy,letscallitthepowerincrement.Forexample,anupgradewithenergyof6hasapowerincrementof1because6isdivisibleby21butnotdivisibleby22ormore,whileforupgradewithenergyof5powerincrementis0,alsoafterbuyingaweaponwithenergyv,Xcanonlybuyupgradeswithenergiesgreaterthanv,inotherwords,theupgradesenergiesthatXisgonnabuyshouldbeinstrictlyincreasingorder.Xiswonderingwhatisthemaximumpowerhecanachieveattheendofhisjourney.notethatonlytheenergyofupgradesshouldbeinstrictlyincreasingordernotthepowerincrementoftheupgrade.1 < n < 100,1 ≤ ai ≤ 106

Input
Thefirstlineofinputcontainsoneinteger,T,whichdenotesnumberoftestcases.Eachtestcasecontainstwolines,thefirstonecontainsoneintegernwhichdenotesnumberofshops,andthenextlinecontainsnintegerstheithofthemdenotestheenergyoftheupgradeprovidedbytheithshop.

Output
PrintoneintegerdenotesmaximumpowerXcanachieveattheendofhisjourney

Example

input

2
3
11016
2
812

output

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810712，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> usingnamespacestd; inta[10000],b[10000]; intdp[10000]; intmain() { intt; cin>>t; while(t--) { memset(dp,0,sizeof(dp)); intn; cin>>n; for(inti=1;i<=n;i++) { intnum=0; intsum=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(inti=1;i<=n;i++) { for(intj=0;j<i;j++) { if(a[i]>a[j]) { dp[i]=max(dp[i],b[i]+dp[j]); } } } intans=0; for(inti=1;i<=n;i++) ans=max(ans,dp[i]); printf("%d\n",ans); } return0; }