QSC and Master 2016沈阳网赛

QSC and Master
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0


Problem Description
Every school has some legends, Northeastern University is the same.

Enter from the north gate of Northeastern University，You are facing the main building of Northeastern University.Ninety-nine percent of the students have not been there，It is said that there is a monster in it.

QSCI am a curious NEU_ACMer，This is the story he told us.

It’s a certain period，QSCI am in a dark night, secretly sneaked into the East Building，hope to see the master.After a serious search，He finally saw the little master in a dark corner. The master said：

“You and I, we're interfacing.please solve my little puzzle！

There are N pairs of numbers，Each pair consists of a key and a value，Now you need to move out some of the pairs to get the score.You can move out two continuous pairs，if and only if their keys are non coprime(their gcd is not one).The final score you get is the sum of all pair’s value which be moved out. May I ask how many points you can get the most？

The answer you give is directly related to your final exam results~The young man~”

QSC is very sad when he told the story，He failed his linear algebra that year because he didn't work out the puzzle.

Could you solve this puzzle?

（Data range：1<=N<=300
1<=Ai.key<=1,000,000,000
0<Ai.value<=1,000,000,000）


Input
First line contains a integer T，means there are T(1≤T≤10) test case。

  Each test case start with one integer N . Next line contains N integers，means Ai.key.Next line contains N integers，means Ai.value.


Output
For each test case,output the max score you could get in a line.


Sample Input

3
3
1 2 3
1 1 1
3
1 2 4
1 1 1
4
1 3 4 3
1 1 1 1




Sample Output

0
2
0

d[i][j]=下标为i到j的子pair
sum[i]=前i个pair的价值的总和
①当i+1 --- j-1的子段全部被消去时 即sum[j-1]-sum[i]==d[i+1][j-1]时
i和j变成相邻的pair对 如果gcd(key[i],key[j])!=0  可以被消去
if(sum[j-1]-sum[i]==d[i-1][j-1]){
    d[i][j]=max(d[i][j],d[i+1][j-1])
    if(gcd(key[i],key[j]!=1)
        d[i][j]=max(d[i][j],d[i+1][j-1]+v[i]+v[j])
}
②显然 i--j的子段 可以被划分为 i -- k 和 k+1 --- j
d[i][j]=max{ d[i][j] , d[i][k]+d[k+1][j] }   i<=k<=j

显然 跑一遍如果任意d[i][j]都没有更新值,再跑几遍也不会改变 而结果也就是d[1][n]了

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
#include<bitset>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007

const int N=305;

ll a[N];
ll v[N];
ll sum[N];
ll d[N][N];

ll gcd(ll a,ll b){
    return b?gcd(b,a%b):a;
}

ll dp(int n){
    for(int i=1;i<=n;++i)
        fill(d[i],d[i]+n+1,0);
    for(int i=1;i<n;++i)
        if(gcd(a[i],a[i+1])!=1)
            d[i][i+1]=v[i]+v[i+1];
    bool update=true;
    while(update){
        update=false;
        for(int i=1;i<=n;++i)
            for(int j=i+2;j<=n;++j){
                if((j-1>i+1)&&(sum[j-1]-sum[i]==d[i+1][j-1])){
                    if(d[i+1][j-1]>d[i][j]){
                        d[i][j]=d[i+1][j-1];
                        update=true;
                    }
                    if((d[i+1][j-1]+v[i]+v[j]>d[i][j])&&(gcd(a[i],a[j])!=1)){
                        d[i][j]=d[i+1][j-1]+v[i]+v[j];
                        update=true;
                    }
                }
                for(int k=i;k<j;++k){
                    if(d[i][k]+d[k+1][j]>d[i][j]){
                        d[i][j]=d[i][k]+d[k+1][j];
                        update=true;
                    }
                }
            }
    }
    return d[1][n];
}

int main()
{
    //freopen("/home/lu/文档/r.txt","r",stdin);
    //freopen("/home/lu/文档/w.txt","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;++i)
            scanf("%lld",a+i);
        for(int i=1;i<=n;++i){
            scanf("%lld",v+i);
            sum[i]=sum[i-1]+v[i];
        }
        cout<<dp(n)<<endl;
    }
    return 0;
}