Codeforces 914D - Bash and a Tough Math Puzzle 線段樹,區間GCD
阿新 • • 發佈:2019-01-06
題意:
兩個操作,
單點修改
詢問一段區間是否能在至多一次修改後,使得區間$GCD$等於$X$
題解:
正確思路;
線段樹維護區間$GCD$,查詢$GCD$的時候記錄一共訪問了多少個$GCD$不被X整除的區間即可,大於一個就NO
要注意的是,如果真的數完一整個區間,肯定會超時,因此用一個外部變數儲存數量,一旦超過一個,就停止整個查詢
#include <bits/stdc++.h> #define endl '\n' #define ll long long #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0) #define rep(ii,a,b) for(int ii=a;ii<=b;++ii) using namespace std; int casn,n,m,k; class segtree{ #define nd node[now] #define ndl node[now<<1] #define ndr node[now<<1|1] public: struct segnode { int l,r;ll gcd,mn; int mid(){return (r+l)>>1;} int len(){return r-l+1;} void update(int x){mn=gcd=x;} }; vector<segnode> node; int cnt; segtree(int n) {node.resize(n<<2|3);maketree(1,n);} void pushup(int now){ nd.gcd=__gcd(ndl.gcd,ndr.gcd); nd.mn=min(ndl.mn,ndr.mn); } void pushdown(int now){} void maketree(int s,int t,int now=1){ nd={s,t,0,0}; if(s==t){ ll x;cin>>x;nd.update(x); return ; } maketree(s,nd.mid(),now<<1); maketree(nd.mid()+1,t,now<<1|1); pushup(now); } void update(int pos,ll x,int now=1){ if(pos>nd.r||pos<nd.l) return ; if(nd.len()==1){nd.update(x);return ;} pushdown(now); update(pos,x,now<<1); update(pos,x,now<<1|1); pushup(now); } int query(int s,int t,ll x){cnt=0;count(s,t,x);return cnt<=1;} void count(int s,int t,ll x,int now=1){ if(cnt>1||s>nd.r||t<nd.l||nd.gcd%x==0) return ; if(nd.len()==1) {cnt++; return ;} count(s,t,x,now<<1);count(s,t,x,now<<1|1); } }; int main() { IO; cin>>n; segtree tree(n); cin>>m; while(m--){ ll a,b,c,d; cin>>a; if(a==1) { cin>>b>>c>>d; if(tree.query(b,c,d)) cout<<"YES"<<endl; else cout<<"NO"<<endl; }else { cin>>b>>c; tree.update(b,c); } } return 0; }
錯誤思路(會WA8):
如果要修改一次使得$GCD$等於$X$,肯定是修改區間的最小值,線段樹維護即可
錯誤原因在於,$GCD$大於$X$的時候,最小值可能是$X$的倍數,此時不應該修改最小值
#include <bits/stdc++.h> #define endl '\n' #define ll long long #define ull unsigned long long #define fi first #define se second #define mp make_pair #define pii pair<int,int> #define all(x) x.begin(),x.end() #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0) #define rep(ii,a,b) for(int ii=a;ii<=b;++ii) #define per(ii,a,b) for(int ii=b;ii>=a;--ii) #define forn(ii,x) for(int ii=head[x];ii;ii=e[ii].next) #pragma GCC optimize("Ofast") #pragma GCC optimize("unroll-loops") #define inline inline __attribute__( \ (always_inline, __gnu_inline__, __artificial__)) \ __attribute__((optimize("Ofast"))) __attribute__((target("sse"))) \ __attribute__((target("sse2"))) __attribute__((target("mmx"))) #define show(x) cout<<#x<<"="<<x<<endl #define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl #define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl #define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl #define show5(v,w,x,y,z) cout<<#v<<" "<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl #define showa(a,b) cout<<#a<<'['<<b<<"]="<<a[b]<<endl using namespace std; const int maxn=1e6+10,maxm=2e6+10; const ll INF=0x3f3f3f3f3f3f; const int mod=1e9+7; const double PI=acos(-1.0); //head int casn,n,m,k; int num[maxn]; class segtree{ #define nd node[now] #define ndl node[now<<1] #define ndr node[now<<1|1] public: struct segnode { int l,r;ll gcd,mn; int mid(){return (r+l)>>1;} int len(){return r-l+1;} void update(int x){mn=gcd=x;} }; vector<segnode> node; segtree(int n) {node.resize(n<<2|3);maketree(1,n);} void pushup(int now){ nd.gcd=__gcd(ndl.gcd,ndr.gcd); nd.mn=min(ndl.mn,ndr.mn); } void pushdown(int now){} void maketree(int s,int t,int now=1){ nd={s,t,0,0}; if(s==t){ ll x;cin>>x;nd.update(x); return ; } maketree(s,nd.mid(),now<<1); maketree(nd.mid()+1,t,now<<1|1); pushup(now); } void update(int pos,ll x,int now=1){ if(pos>nd.r||pos<nd.l) return ; if(nd.len()==1){nd.update(x);return ;} pushdown(now); update(pos,x,now<<1); update(pos,x,now<<1|1); pushup(now); } ll query_minid(int s,int t,int now=1){ if(nd.len()==1) return s; if(ndl.mn<=ndr.mn) return query_minid(s,t,now<<1); else return query_minid(s,t,now<<1|1); } ll query_min(int s,int t,int now=1){ if(s>nd.r||t<nd.l) return INF; if(s<=nd.l&&nd.r<=t) return nd.mn; return min(query_min(s,t,now<<1),query_min(s,t,now<<1|1)); } ll query_gcd(int s,int t,int now=1){ if(s>nd.r||t<nd.l) return 0; if(s<=nd.l&&t>=nd.r)return nd.gcd; return __gcd(query_gcd(s,t,now<<1),query_gcd(s,t,now<<1|1)); } }; int main() { //#define test #ifdef test auto _start = chrono::high_resolution_clock::now(); freopen("in.txt","r",stdin);freopen("out.txt","w",stdout); #endif // IO; cin>>n; segtree tree(n); cin>>m; while(m--){ ll a,b,c,d; cin>>a; if(a==1) { cin>>b>>c>>d; int id=tree.query_minid(b,c); ll mn=tree.query_min(b,c); tree.update(id,d); if(tree.query_gcd(b,c)==d)cout<<"YES"<<endl; else cout<<"NO"<<endl; tree.update(id,mn); }else { cin>>b>>c; tree.update(b,c); } } return 0; }