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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
#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|1 const 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; }
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原文地址:http://www.cnblogs.com/dyzll/p/5882738.html