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2016沈阳网络赛 QSC and Master
QSC and Master
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
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）
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.
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
331 2 31 1 131 2 41 1 141 3 4 31 1 1 1
Sample Output
020
分析：dp转移时要么取断开两部分的值的最大和，要么是两边都能取；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int n,m,k,t;ll dp[maxn][maxn],a[maxn],b[maxn],sum[maxn];int main(){ int i,j; scanf("%d",&t); while(t--) { scanf("%d",&n); memset(dp,0,sizeof dp); rep(i,1,n)scanf("%lld",&a[i]); rep(i,1,n)scanf("%lld",&b[i]),sum[i]=sum[i-1]+b[i]; rep(i,2,n) { for(j=1;j+i-1<=n;j++) { if(gcd(a[j],a[j+i-1])!=1&&dp[j+1][i-2]==sum[j+i-2]-sum[j])dp[j][i]=b[j]+b[j+i-1]+dp[j+1][i-2]; for(k=j+1;k<=j+i-1;k++)dp[j][i]=max(dp[j][i],dp[j][k-j]+dp[k][j+i-k]); } } printf("%lld\n",dp[1][n]); } //system("Pause"); return 0;}
2016沈阳网络赛 QSC and Master
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