BZOJ 4868-4873 题解

BZOJ4868

每个结束位置的最优值很显然具有单调性,三分,再讨论一下就好了.

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define FILE "exam"
 #define up(i,j,n) for(int i=j;i<=n;i++)
 #define db long double
 #define pii pair<int,int>
 #define pb push_back
 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
 template<class T> inline T squ(T a){return a*a;}
 const int maxn=+,inf=1e9+,mod=;
 ll read(){
     ll x=,f=,ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<='')x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x*f;
 }
 db A,B,C;
 ll a[maxn],b[maxn],d[maxn],n,m;
 db check(ll s){
     db ans=;
     memcpy(d,b,sizeof(d));
     if(A<B){
         int l=,r=m;
         while(l<r){
             if(d[r]<s||d[l]>s)break;
             if(d[r]-s>s-d[l])ans+=A*(s-d[l]),d[r]-=s-d[l],d[l]=s,l++;
             else ans+=A*(d[r]-s),d[l]+=d[r]-s,d[r]=s,r--;
         }
         for(int i=m;i>=;i--)
             if(d[i]>s)ans+=B*(d[i]-s);
         for(int i=;i<=n;i++)
             if(a[i]<s)ans+=C*(s-a[i]);
         return ans;
     }
     else {
         for(int i=;i<=m;i++)if(d[i]>s)ans+=B*(d[i]-s);
         for(int i=;i<=n;i++)
             if(a[i]<s)ans+=C*(s-a[i]);
         return ans;
     }
 }
 int main(){
     freopen(FILE".in","r",stdin);
     freopen(FILE".out","w",stdout);
     A=read(),B=read(),C=read();
     n=read(),m=read();
     ll left=inf,right=-inf;
     up(i,,n)a[i]=read(),cmin(left,a[i]);
     up(i,,m)b[i]=read(),cmax(right,b[i]);
     sort(b+,b+m+);
     sort(a+,a+n+);
     while(right-left>){
         int mid1=(left+left+right)/;
         int mid2=(left+right+right)/;
         if(check(mid1)>check(mid2))left=mid1;
         else right=mid2;
     }
     db Ans=(ll)1e18;
     for(int i=left;i<=right;i++)
         cmin(Ans,check(i));
     printf("%.0Lf\n",Ans);
     return ;
 }

BZOJ4869

看到这道题后我想到了某道同样是一堆幂的神题,尽管我没写过...

正确做法是EX欧拉定理+线段树,我会欧拉定理,但我不知道什么是EX欧拉定理.

请自行百度,(尽管你baidu不到).

 #include<bits/stdc++.h>
 using namespace std;
 #define int long long
 #define FILE "verbinden"
 #define up(i,j,n) for(int i=j;i<=n;++i)
 #define db long double
 #define pii pair<int,int>
 #define pb push_back
 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
 template<class T> inline T squ(T a){return a*a;}
 const int maxn=+,inf=1e9+;
 int read(){
     int x=,f=,ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<='')x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x*f;
 }
 int n,m,mod,C;
 int a[maxn];
 int op[maxn],x[maxn],y[maxn];
 int fi[maxn];
 int getfi(int n){
     int m=sqrt(n*1.0)+,ans=;
     if(n==)return ;
     for(int i=;i<=m;i++){
         if(n%i==)ans=ans*(i-),n/=i;
         while(n%i==){
             ans=ans*(i);
             n/=i;
         }
         if(n==)break;
     }
     if(n!=)ans=ans*(n-);
     return ans;
 }
 int sum[maxn],siz[maxn],ci[maxn],tot=;
 void updata(int o){
     sum[o]=(sum[o<<]+sum[o<<|])%mod;
     siz[o]=siz[o<<]+siz[o<<|];
 }
 void build(int l,int r,int o){
     if(l==r){
         sum[o]=a[l];
         siz[o]=;
         return;
     }
     int mid=(l+r)>>;
     build(l,mid,o<<);
     build(mid+,r,o<<|);
     updata(o);
 }
 bool flag=;
 int qpow(int a,int b,int mod){
     int ans=;
     bool f0=;
     while(b){
         if(b&){
             if(ans*a>=mod||f0)flag=;
             ans=ans*a%mod;
         }
         if(squ(a)>=mod)f0=;
         a=squ(a)%mod;
         b>>=;
     }
     return ans;
 }
 int k(int a,int mod){
     if(a>=mod)return a%mod+mod;
     else return a;
 }
 void change(int l,int r,int L,int R,int o){
     if(l>R||r<L)return ;
     if(l==r){
         if(siz[o]){
             ci[l]++;
             sum[o]=k(a[l],fi[ci[l]]);
             for(int j=ci[l]-;j>=;j--){
                 flag=;
                 sum[o]=qpow(C,sum[o],fi[j]);
                 if(flag)sum[o]+=fi[j];
             }
             if(ci[l]==tot)siz[o]=;
         }
         return;
     }
     int mid=(l+r)>>;
     if(l>=L&&r<=R){
         if(siz[o<<])change(l,mid,L,R,o<<);
         if(siz[o<<|])change(mid+,r,L,R,o<<|);
         updata(o);
         return;
     }
     change(l,mid,L,R,o<<);
     change(mid+,r,L,R,o<<|);
     updata(o);
 }
 int query(int l,int r,int L,int R,int o){
     if(l>R||r<L)return ;
     if(l>=L&&r<=R)return sum[o];
     int mid=(l+r)>>;
     return (query(l,mid,L,R,o<<)+query(mid+,r,L,R,o<<|))%mod;
 }
 main(){
     freopen(FILE".in","r",stdin);
     freopen(FILE".out","w",stdout);
     n=read(),m=read(),mod=read(),C=read();
     up(i,,n)a[i]=read();
     up(i,,m)op[i]=read(),x[i]=read(),y[i]=read();
     build(,n,);
     fi[]=mod;
     for(int i=;i<=m;i++){
         fi[i]=getfi(fi[i-]);
         if(fi[i]==){
             fi[i+]=;
             tot=i+;
             break;
         }
     }
     up(i,,m){
         if(op[i]==)change(,n,x[i],y[i],);
         if(op[i]==)printf("%d\n",query(,n,x[i],y[i],));
     }
     return ;
 }

BZOJ4870

搞出DP方程,然后DP快速幂,或者矩阵快速幂任你挑.(这题可以搞NTT优化).

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define FILE "problem"
 #define up(i,j,n) for(int i=j;i<=n;i++)
 #define db long double
 #define pii pair<int,int>
 #define pb push_back
 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
 template<class T> inline T squ(T a){return a*a;}
 const int maxn=+,inf=1e9+;
 int read(){
     int x=,f=,ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<='')x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x*f;
 }
 ll mod,n,K,r;
 void getmod(ll& a){if(a>mod)a-=mod;}
 struct Matrix{
     ll num[][];
     Matrix(){memset(num,,sizeof(num));}
     Matrix(ll a){
         memset(num,,sizeof(num));
         for(int i=;i<K;i++)
             num[i][i]=a;
     }
     Matrix operator*(const Matrix& b){
         Matrix c;
         for(int i=;i<K;i++)
             for(int j=;j<K;j++)
                 for(int k=;k<K;k++)
                     getmod(c.num[i][j]+=num[i][k]*b.num[k][j]%mod);
         return c;
     }
 }b,a;

 int main(){
     freopen(FILE".in","r",stdin);
     freopen(FILE".out","w",stdout);
     n=read(),mod=read(),K=read(),r=read();
     for(int i=;i<K;i++)
         b.num[i][(i+)%K]++,b.num[i][i]++;
     Matrix ans();
     n=n*K;
     while(n){
         if(n&)ans=ans*b;
         b=b*b;
         n>>=;
     }
     printf("%lld\n",ans.num[][r]);
     return ;
 }

BZOJ 4871

一道状态有些复杂的树形DP.

状态自行baidu.

注:此方法已被HACK,请自行找不会被卡的方法.

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 #define ll long long
 #define FILE "treediagram"
 #define up(i,j,n) for(int i=j;i<=n;i++)
 #define db long double
 #define pii pair<int,int>
 #define pb push_back
 #define mem(a,L) memset(a,0,sizeof(int)*(L+1))
 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
 template<class T> inline T squ(T a){return a*a;}
 const int maxn=+,inf=1e9+,mod=;
 int read(){
     int x=,f=,ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<='')x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x*f;
 }
 struct node{
     int x,y,next;
 }e[maxn];
 int len,linkk[maxn],T,ch,p0,h0,p1,h1,n;
 int vis[maxn],de[maxn],h[maxn],f[maxn],g[maxn],d[maxn];
 void insert(int x,int y){e[++len].y=y;e[len].x=x;e[len].next=linkk[x];linkk[x]=len;}
 struct Node{
     int a,b;
     Node(){a=b=;}
 };
 void dp(int p){
     if(vis[p])return;vis[p]=;
     int x=e[p].y,child=;
     Node t;
     for(int i=linkk[x];i;i=e[i].next) if(i!=(p^)){
         child++;
         dp(i);
         cmax(d[p],d[i]);
         if(f[i]>=t.a)t.b=t.a,t.a=f[i];
         else if(f[i]>t.b)t.b=f[i];
     }
     f[p]=max(t.a,)+child-;
     g[p]=max(t.a,)+max(t.b,)+child-;
     cmax(d[p],g[p]);
 }
 int a[],b[];
 bool cmp(const int& a,const int& b){return a>b;}
 void solve(){
     int ans=;
     for(int i=;i<=len;i++)dp(i),dp(i^),cmax(ans,d[i]+d[i^]);
     for(int x=;x<=n;x++){
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         for(int i=linkk[x];i;i=e[i].next){
             a[]=f[i];sort(a,a+,cmp);
             b[]=d[i];sort(b,b+,cmp);
         }
         cmax(ans,b[]+b[]+);
         cmax(ans,a[]+a[]+a[]+de[x]-);
         cmax(ans,a[]+a[]+a[]+a[]+de[x]-);
     }
     printf("%d\n",ans);
 }
 int main(){
     freopen(FILE".in","r",stdin);
     freopen(FILE".out","w",stdout);
     int T=read(),ch=read();
     while(T--){
         n=read();
         if(ch)p0=read(),p1=read();
         if(ch>)h0=read(),h1=read();
         if(n==){puts("");continue;}
         len=;mem(linkk,n);mem(de,n);
         up(i,,n){int x=read(),y=read();insert(x,y);insert(y,x);de[x]++,de[y]++;}
         up(i,,len)vis[i]=,f[i]=g[i]=d[i]=;
         solve();
     }
     return ;
 }

BZOJ 4872

概率DP.

给个我能理解的讲解的链接:http://www.cnblogs.com/GXZlegend/p/6764969.html

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define FILE "trennen"
 #define up(i,j,n) for(ll i=j;i<=n;i++)
 #define db long double
 #define pii pair<ll,ll>
 #define pb push_back
 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
 template<class T> inline T squ(T a){return a*a;}
 const ll maxn=+,inf=1e9+,mod=;
 ll read(){
     ll x=,f=,ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<='')x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x*f;
 }
 ll n,k;
 ll a[maxn],ans=;
 ll fac[maxn];
 ll qpow(ll a,ll b){
     ll ans=;
     while(b){
         if(b&)ans=ans*a%mod;
         a=squ(a)%mod;
         b>>=;
     }
     return ans;
 }
 ll vis[maxn];
 int main(){
     freopen(FILE".in","r",stdin);
     freopen(FILE".out","w",stdout);
     n=read(),k=read();
     up(i,,n)a[i]=read();
     for(ll i=n;i>=;i--){
         ll x=a[i];
         for(ll j=;j*i<=n;j++)
             x^=vis[j*i];
         if(x)ans++,vis[i]=;
     }
     fac[]=;for(ll i=;i<=n;i++)fac[i]=fac[i-]*i%mod;
     printf("%lld\n",(ll)ans*fac[n]%mod);
     return ;
 }

BZOJ 4873

最简单的那种最大权闭合子图.

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 #define ll long long
 #define FILE "sushi"
 #define up(i,j,n) for(int i=j;i<=n;i++)
 #define db long double
 #define pii pair<int,int>
 #define pb push_back
 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
 template<class T> inline T squ(T a){return a*a;}
 const int maxn=+,inf=1e8+,mod=;
 int read(){
     int x=,f=,ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<='')x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x*f;
 }
 int S,T,n,m,cnt;
 struct node{
     int y,next,flow,rev;
     node(int y=,int next=,int flow=,int rev=):y(y),next(next),flow(flow),rev(rev){}
 }e[maxn];int len,linkk[maxn];
 int v[][];
 void insert(int x,int y,int flow){
     e[++len].y=y;e[len].next=linkk[x];linkk[x]=len;e[len].flow=flow;e[len].rev=len+;
     e[++len].y=x;e[len].next=linkk[y];linkk[y]=len;e[len].flow=;e[len].rev=len-;
 }
 int q[maxn],d[maxn],head,tail;
 bool makelevel(){
     memset(d,-,sizeof(int)*(cnt+));
     d[S]=;head=tail=;q[++tail]=S;
     while(++head<=tail){
         int x=q[head];
         for(int i=linkk[x];i;i=e[i].next){
             int y=e[i].y;
             if(d[y]==-&&e[i].flow)
                 q[++tail]=y,d[y]=d[x]+;
         }
     }
     return d[T]!=-;
 }
 int makeflow(int x,int flow){
     if(x==T||!flow)return flow;
     int maxflow=,dis=;
     for(int i=linkk[x];i&&maxflow<flow;i=e[i].next){
         int y=e[i].y;
         if(d[y]==d[x]+&&e[i].flow)
             if(dis=makeflow(y,min(flow-maxflow,e[i].flow))){
                 e[i].flow-=dis;
                 e[e[i].rev].flow+=dis;
                 maxflow+=dis;
             }
     }
     if(!maxflow)d[x]=-;
     return maxflow;
 }
 int dinic(){
     int ans=,d;
     while(makelevel())
         while(d=makeflow(S,inf))
             ans+=d;
     return ans;
 }
 int ID1[][],ID2[maxn],a[maxn];
 int main(){
     freopen(FILE".in","r",stdin);
     freopen(FILE".out","w",stdout);
     n=read();m=read();
     up(i,,n)a[i]=read();
     up(i,,n)up(j,i,n){
         v[i][j]=read();if(i==j)v[i][j]-=a[i];
         ID1[i][j]=++cnt;
     }
     up(i,,n)up(j,i,n)if(i!=j)insert(ID1[i][j],ID1[i][j-],inf),insert(ID1[i][j],ID1[i+][j],inf);
     S=++cnt,T=++cnt;
     if(m==)up(i,,n){
         if(!ID2[a[i]]){
             ID2[a[i]]=++cnt;
             insert(ID2[a[i]],T,squ(a[i]));
         }
         insert(ID1[i][i],ID2[a[i]],inf);
     }
     int Sum=;
     up(i,,n)up(j,i,n){
         if(v[i][j]>)insert(S,ID1[i][j],v[i][j]),Sum+=v[i][j];
         else if(v[i][j]<)insert(ID1[i][j],T,-v[i][j]);
     }
     int ans=dinic();
     printf("%d\n",Sum-ans);
     return ;
 }

我们用这套题考了两天.

我的考试结果并不是很理想.

考完发现全是傻逼题.

发现了两点很不好的地方是:

1.考试时不敢想正解.(出题人语文水平高超)

2.很容易被前面的引导.(容易被出题人引导)　

考试时题面的长度不一定与难度成反比.

很多题目打暴力的难度比打正解大.

因为正解的模型我可能建过很多次,但暴力的模型我第一次建.

但有些时候打暴力比打正解划算,因为你不会正解...

所以说了这么多,你需要做的是:

考试时摆脱前面题目不会的心理负担,全力分析这道题,争取AC,当然不能AC就只能暴力了.

蛤蛤,其实我说了一堆废话.