QSC and Master

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2408 Accepted Submission(s): 876

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

Source

2016 ACM/ICPC Asia Regional Shenyang Online

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=310;
const int INF=0x3f3f3f3f;
typedef long long ll;
ll dp[maxn][maxn], sum[maxn],  val[maxn];
int id[maxn];
int gcd(int x, int y){
    return y==0 ? x : gcd(y, x%y);
}
int main(){
    int T, n;
    scanf("%d", &T);
    while(T--){
        memset(dp, 0, sizeof(dp));
        memset(sum, 0, sizeof(sum));
        scanf("%d", &n);
        for(int i=1; i<=n; i++)
            scanf("%d", &id[i]);
        for(int j=1; j<=n; j++)
        {
             scanf("%lld", &val[j]);
             sum[j]=sum[j-1]+val[j];
        }
        for(int l=1; l<n; l++)
        {
            for(int i=1; i+l<=n; i++)//枚举区间
            {
                int j=i+l;
                for(int k=i; k<j; k++)
                    dp[i][j]=max(dp[i][j], dp[i][k]+dp[k+1][j]);//更新此区间
                if(gcd(id[i], id[j])>1)//将满足条件的区间计算出来
                {
                    if(i+1==j)
                        dp[i][j]=sum[j]-sum[i-1];
                    else if(dp[i+1][j-1]==sum[j-1]-sum[i])
                        dp[i][j]=sum[j]-sum[i-1];
                }
            }
        }
        printf("%lld\n", dp[1][n]);
    }
    return 0;
}