题目链接
CF402D Upgrading Array (luogu)
CF402D Upgrading Array
解题思路
首先有一个很显然的结论,就是你一旦在第 \(i\) 个位置上做了一次操作后,那么你之后所有在第 \(j(i \le j)\) 个位置做的操作都无效了,因为此时前缀的公因数为 \(1\) 了。
因此有个很显然的结论就是操作需要从后往前做。
然后我们只做对有贡献的操作,有贡献指的是做了这个操作后答案大于等于原本的答案或者说等价于这个操作的区间的最大公因数的价值小于 \(0\)。
于是根据以上方式进行操作即可,时间复杂度 \(O(n \sqrt v )\)。
不过值得注意的是,本题比较卡常,记得在各个地方加上记忆化以加快程序效率。
参考代码
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#include<bits/stdc++.h>
using namespace std;
#define map unordered_map
#define re register
#define ll long long
#define forl(i,a,b) for(re ll i=a;i<=b;i++)
#define forr(i,a,b) for(re ll i=a;i>=b;i--)
#define forll(i,a,b,c) for(re ll i=a;i<=b;i+=c)
#define forrr(i,a,b,c) for(re ll i=a;i>=b;i-=c)
#define lc(x) x<<1
#define rc(x) x<<1|1
#define mid ((l+r)>>1)
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) (x&-x)
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define db long double
#define ull unsigned long long
#define lcm(x,y) x/__gcd(x,y)*y
#define Sum(x,y) 1ll*(x+y)*(y-x+1)/2
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
#define maxqueue priority_queue<ll>
#define minqueue priority_queue<ll,vector<ll>,greater<ll>>
ll t;
ll n,m,ans=-1e18;
map<ll,ll>mp,mp2;
ll a[100010],b[100010],lst[100010];
ll lst2;
ll gcd[100010];
map<ll,ll>mp3;
bool pdzs(ll a)
{if(mp3[a])return mp3[a]-1;if(a==1)return 0;if(a==2)return 1;for(ll i=2;i<=sqrt(a);i++)if(a%i==0){mp3[a]=1;return 0;}mp3[a]=2;return 1;
}
bool pd[1000010];
long long A[2000010],K;
void work(ll n)
{for(ll i=2;i<=n;i++){if(pd[i]==0)A[++K]=i;for(ll j=1;j<=K && i*A[j]<=n;j++){pd[i*A[j]]=1;if(i%A[j]==0)break;}}
}
ll f(ll x)
{ll num=x;if(mp2[num])return mp2[num];ll an=0;if(pdzs(num))return mp[num]==0?1:-1;forl(i,1,K)if(x%A[i]==0){ll pd=mp[A[i]]==0?1:-1;while(x%A[i]==0)x/=A[i],an+=pd;if(x==1)return mp2[num]=an;}if(pdzs(x))return mp2[num]=an+(mp[x]==0?1:-1);return mp2[num]=an;
}
void solve()
{lst2=1;cin>>n>>m;forl(i,1,n)cin>>a[i],lst[i]=a[i];gcd[1]=a[1];forl(i,2,n)gcd[i]=__gcd(gcd[i-1],a[i]);forl(i,1,m)cin>>b[i],mp[b[i]]=1;ll sum=0;forl(i,1,n)sum+=f(a[i]);
// cout<<sum<<endl;forr(i,n,1)if(f(gcd[i]/lst2)<0)sum+=f(gcd[i]/lst2)*-1*i,lst2=gcd[i];cout<<max(ans,sum)<<endl;
}
int main()
{work(40000);IOS;t=1;// cin>>t;while(t--)solve();QwQ;
}