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.
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define FILE "exam" 5 #define up(i,j,n) for(int i=j;i<=n;i++) 6 #define db long double 7 #define pii pair<int,int> 8 #define pb push_back 9 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 10 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 11 template<class T> inline T squ(T a){return a*a;} 12 const int maxn=105000+10,inf=1e9+10,mod=201314; 13 ll read(){ 14 ll x=0,f=1,ch=getchar(); 15 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 16 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 17 return x*f; 18 } 19 db A,B,C; 20 ll a[maxn],b[maxn],d[maxn],n,m; 21 db check(ll s){ 22 db ans=0; 23 memcpy(d,b,sizeof(d)); 24 if(A<B){ 25 int l=1,r=m; 26 while(l<r){ 27 if(d[r]<s||d[l]>s)break; 28 if(d[r]-s>s-d[l])ans+=A*(s-d[l]),d[r]-=s-d[l],d[l]=s,l++; 29 else ans+=A*(d[r]-s),d[l]+=d[r]-s,d[r]=s,r--; 30 } 31 for(int i=m;i>=1;i--) 32 if(d[i]>s)ans+=B*(d[i]-s); 33 for(int i=1;i<=n;i++) 34 if(a[i]<s)ans+=C*(s-a[i]); 35 return ans; 36 } 37 else { 38 for(int i=1;i<=m;i++)if(d[i]>s)ans+=B*(d[i]-s); 39 for(int i=1;i<=n;i++) 40 if(a[i]<s)ans+=C*(s-a[i]); 41 return ans; 42 } 43 } 44 int main(){ 45 freopen(FILE".in","r",stdin); 46 freopen(FILE".out","w",stdout); 47 A=read(),B=read(),C=read(); 48 n=read(),m=read(); 49 ll left=inf,right=-inf; 50 up(i,1,n)a[i]=read(),cmin(left,a[i]); 51 up(i,1,m)b[i]=read(),cmax(right,b[i]); 52 sort(b+1,b+m+1); 53 sort(a+1,a+n+1); 54 while(right-left>3){ 55 int mid1=(left+left+right)/3; 56 int mid2=(left+right+right)/3; 57 if(check(mid1)>check(mid2))left=mid1; 58 else right=mid2; 59 } 60 db Ans=(ll)1e18; 61 for(int i=left;i<=right;i++) 62 cmin(Ans,check(i)); 63 printf("%.0Lf\n",Ans); 64 return 0; 65 }
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).
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define int long long 4 #define FILE "verbinden" 5 #define up(i,j,n) for(int i=j;i<=n;++i) 6 #define db long double 7 #define pii pair<int,int> 8 #define pb push_back 9 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 10 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 11 template<class T> inline T squ(T a){return a*a;} 12 const int maxn=210000+10,inf=1e9+10; 13 int read(){ 14 int x=0,f=1,ch=getchar(); 15 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 16 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 17 return x*f; 18 } 19 int n,m,mod,C; 20 int a[maxn]; 21 int op[maxn],x[maxn],y[maxn]; 22 int fi[maxn]; 23 int getfi(int n){ 24 int m=sqrt(n*1.0)+1,ans=1; 25 if(n==1)return 1; 26 for(int i=2;i<=m;i++){ 27 if(n%i==0)ans=ans*(i-1),n/=i; 28 while(n%i==0){ 29 ans=ans*(i); 30 n/=i; 31 } 32 if(n==1)break; 33 } 34 if(n!=1)ans=ans*(n-1); 35 return ans; 36 } 37 int sum[maxn],siz[maxn],ci[maxn],tot=0; 38 void updata(int o){ 39 sum[o]=(sum[o<<1]+sum[o<<1|1])%mod; 40 siz[o]=siz[o<<1]+siz[o<<1|1]; 41 } 42 void build(int l,int r,int o){ 43 if(l==r){ 44 sum[o]=a[l]; 45 siz[o]=1; 46 return; 47 } 48 int mid=(l+r)>>1; 49 build(l,mid,o<<1); 50 build(mid+1,r,o<<1|1); 51 updata(o); 52 } 53 bool flag=0; 54 int qpow(int a,int b,int mod){ 55 int ans=1; 56 bool f0=0; 57 while(b){ 58 if(b&1){ 59 if(ans*a>=mod||f0)flag=1; 60 ans=ans*a%mod; 61 } 62 if(squ(a)>=mod)f0=1; 63 a=squ(a)%mod; 64 b>>=1; 65 } 66 return ans; 67 } 68 int k(int a,int mod){ 69 if(a>=mod)return a%mod+mod; 70 else return a; 71 } 72 void change(int l,int r,int L,int R,int o){ 73 if(l>R||r<L)return ; 74 if(l==r){ 75 if(siz[o]){ 76 ci[l]++; 77 sum[o]=k(a[l],fi[ci[l]]); 78 for(int j=ci[l]-1;j>=0;j--){ 79 flag=0; 80 sum[o]=qpow(C,sum[o],fi[j]); 81 if(flag)sum[o]+=fi[j]; 82 } 83 if(ci[l]==tot)siz[o]=0; 84 } 85 return; 86 } 87 int mid=(l+r)>>1; 88 if(l>=L&&r<=R){ 89 if(siz[o<<1])change(l,mid,L,R,o<<1); 90 if(siz[o<<1|1])change(mid+1,r,L,R,o<<1|1); 91 updata(o); 92 return; 93 } 94 change(l,mid,L,R,o<<1); 95 change(mid+1,r,L,R,o<<1|1); 96 updata(o); 97 } 98 int query(int l,int r,int L,int R,int o){ 99 if(l>R||r<L)return 0; 100 if(l>=L&&r<=R)return sum[o]; 101 int mid=(l+r)>>1; 102 return (query(l,mid,L,R,o<<1)+query(mid+1,r,L,R,o<<1|1))%mod; 103 } 104 main(){ 105 freopen(FILE".in","r",stdin); 106 freopen(FILE".out","w",stdout); 107 n=read(),m=read(),mod=read(),C=read(); 108 up(i,1,n)a[i]=read(); 109 up(i,1,m)op[i]=read(),x[i]=read(),y[i]=read(); 110 build(1,n,1); 111 fi[0]=mod; 112 for(int i=1;i<=m;i++){ 113 fi[i]=getfi(fi[i-1]); 114 if(fi[i]==1){ 115 fi[i+1]=1; 116 tot=i+1; 117 break; 118 } 119 } 120 up(i,1,m){ 121 if(op[i]==0)change(1,n,x[i],y[i],1); 122 if(op[i]==1)printf("%d\n",query(1,n,x[i],y[i],1)); 123 } 124 return 0; 125 }
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).
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define FILE "problem" 5 #define up(i,j,n) for(int i=j;i<=n;i++) 6 #define db long double 7 #define pii pair<int,int> 8 #define pb push_back 9 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 10 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 11 template<class T> inline T squ(T a){return a*a;} 12 const int maxn=110000+10,inf=1e9+10; 13 int read(){ 14 int x=0,f=1,ch=getchar(); 15 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 16 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 17 return x*f; 18 } 19 ll mod,n,K,r; 20 void getmod(ll& a){if(a>mod)a-=mod;} 21 struct Matrix{ 22 ll num[52][52]; 23 Matrix(){memset(num,0,sizeof(num));} 24 Matrix(ll a){ 25 memset(num,0,sizeof(num)); 26 for(int i=0;i<K;i++) 27 num[i][i]=a; 28 } 29 Matrix operator*(const Matrix& b){ 30 Matrix c; 31 for(int i=0;i<K;i++) 32 for(int j=0;j<K;j++) 33 for(int k=0;k<K;k++) 34 getmod(c.num[i][j]+=num[i][k]*b.num[k][j]%mod); 35 return c; 36 } 37 }b,a; 38 39 int main(){ 40 freopen(FILE".in","r",stdin); 41 freopen(FILE".out","w",stdout); 42 n=read(),mod=read(),K=read(),r=read(); 43 for(int i=0;i<K;i++) 44 b.num[i][(i+1)%K]++,b.num[i][i]++; 45 Matrix ans(1); 46 n=n*K; 47 while(n){ 48 if(n&1)ans=ans*b; 49 b=b*b; 50 n>>=1; 51 } 52 printf("%lld\n",ans.num[0][r]); 53 return 0; 54 }
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.
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 #define ll long long 6 #define FILE "treediagram" 7 #define up(i,j,n) for(int i=j;i<=n;i++) 8 #define db long double 9 #define pii pair<int,int> 10 #define pb push_back 11 #define mem(a,L) memset(a,0,sizeof(int)*(L+1)) 12 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 13 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 14 template<class T> inline T squ(T a){return a*a;} 15 const int maxn=210000+10,inf=1e9+10,mod=10003; 16 int read(){ 17 int x=0,f=1,ch=getchar(); 18 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 19 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 20 return x*f; 21 } 22 struct node{ 23 int x,y,next; 24 }e[maxn]; 25 int len,linkk[maxn],T,ch,p0,h0,p1,h1,n; 26 int vis[maxn],de[maxn],h[maxn],f[maxn],g[maxn],d[maxn]; 27 void insert(int x,int y){e[++len].y=y;e[len].x=x;e[len].next=linkk[x];linkk[x]=len;} 28 struct Node{ 29 int a,b; 30 Node(){a=b=0;} 31 }; 32 void dp(int p){ 33 if(vis[p])return;vis[p]=1; 34 int x=e[p].y,child=0; 35 Node t; 36 for(int i=linkk[x];i;i=e[i].next) if(i!=(p^1)){ 37 child++; 38 dp(i); 39 cmax(d[p],d[i]); 40 if(f[i]>=t.a)t.b=t.a,t.a=f[i]; 41 else if(f[i]>t.b)t.b=f[i]; 42 } 43 f[p]=max(t.a,1)+child-1; 44 g[p]=max(t.a,1)+max(t.b,1)+child-2; 45 cmax(d[p],g[p]); 46 } 47 int a[5],b[3]; 48 bool cmp(const int& a,const int& b){return a>b;} 49 void solve(){ 50 int ans=0; 51 for(int i=2;i<=len;i++)dp(i),dp(i^1),cmax(ans,d[i]+d[i^1]); 52 for(int x=1;x<=n;x++){ 53 memset(a,0,sizeof(a)); 54 memset(b,0,sizeof(b)); 55 for(int i=linkk[x];i;i=e[i].next){ 56 a[4]=f[i];sort(a,a+5,cmp); 57 b[2]=d[i];sort(b,b+3,cmp); 58 } 59 cmax(ans,b[0]+b[1]+1); 60 cmax(ans,a[0]+a[1]+a[2]+de[x]-3); 61 cmax(ans,a[0]+a[1]+a[2]+a[3]+de[x]-4); 62 } 63 printf("%d\n",ans); 64 } 65 int main(){ 66 freopen(FILE".in","r",stdin); 67 freopen(FILE".out","w",stdout); 68 int T=read(),ch=read(); 69 while(T--){ 70 n=read(); 71 if(ch)p0=read(),p1=read(); 72 if(ch>1)h0=read(),h1=read(); 73 if(n==1){puts("0");continue;} 74 len=1;mem(linkk,n);mem(de,n); 75 up(i,2,n){int x=read(),y=read();insert(x,y);insert(y,x);de[x]++,de[y]++;} 76 up(i,2,len)vis[i]=0,f[i]=g[i]=d[i]=0; 77 solve(); 78 } 79 return 0; 80 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define FILE "trennen" 5 #define up(i,j,n) for(ll i=j;i<=n;i++) 6 #define db long double 7 #define pii pair<ll,ll> 8 #define pb push_back 9 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 10 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 11 template<class T> inline T squ(T a){return a*a;} 12 const ll maxn=210000+10,inf=1e9+10,mod=100003; 13 ll read(){ 14 ll x=0,f=1,ch=getchar(); 15 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 16 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 17 return x*f; 18 } 19 ll n,k; 20 ll a[maxn],ans=0; 21 ll fac[maxn]; 22 ll qpow(ll a,ll b){ 23 ll ans=1; 24 while(b){ 25 if(b&1)ans=ans*a%mod; 26 a=squ(a)%mod; 27 b>>=1; 28 } 29 return ans; 30 } 31 ll vis[maxn]; 32 int main(){ 33 freopen(FILE".in","r",stdin); 34 freopen(FILE".out","w",stdout); 35 n=read(),k=read(); 36 up(i,1,n)a[i]=read(); 37 for(ll i=n;i>=1;i--){ 38 ll x=a[i]; 39 for(ll j=2;j*i<=n;j++) 40 x^=vis[j*i]; 41 if(x)ans++,vis[i]=1; 42 } 43 fac[0]=1;for(ll i=1;i<=n;i++)fac[i]=fac[i-1]*i%mod; 44 printf("%lld\n",(ll)ans*fac[n]%mod); 45 return 0; 46 }
BZOJ 4873
最简单的那种最大权闭合子图.
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 #define ll long long 6 #define FILE "sushi" 7 #define up(i,j,n) for(int i=j;i<=n;i++) 8 #define db long double 9 #define pii pair<int,int> 10 #define pb push_back 11 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 12 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 13 template<class T> inline T squ(T a){return a*a;} 14 const int maxn=510000+10,inf=1e8+10,mod=100003; 15 int read(){ 16 int x=0,f=1,ch=getchar(); 17 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 18 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 19 return x*f; 20 } 21 int S,T,n,m,cnt; 22 struct node{ 23 int y,next,flow,rev; 24 node(int y=0,int next=0,int flow=0,int rev=0):y(y),next(next),flow(flow),rev(rev){} 25 }e[maxn];int len,linkk[maxn]; 26 int v[330][330]; 27 void insert(int x,int y,int flow){ 28 e[++len].y=y;e[len].next=linkk[x];linkk[x]=len;e[len].flow=flow;e[len].rev=len+1; 29 e[++len].y=x;e[len].next=linkk[y];linkk[y]=len;e[len].flow=0;e[len].rev=len-1; 30 } 31 int q[maxn],d[maxn],head,tail; 32 bool makelevel(){ 33 memset(d,-1,sizeof(int)*(cnt+1)); 34 d[S]=0;head=tail=0;q[++tail]=S; 35 while(++head<=tail){ 36 int x=q[head]; 37 for(int i=linkk[x];i;i=e[i].next){ 38 int y=e[i].y; 39 if(d[y]==-1&&e[i].flow) 40 q[++tail]=y,d[y]=d[x]+1; 41 } 42 } 43 return d[T]!=-1; 44 } 45 int makeflow(int x,int flow){ 46 if(x==T||!flow)return flow; 47 int maxflow=0,dis=0; 48 for(int i=linkk[x];i&&maxflow<flow;i=e[i].next){ 49 int y=e[i].y; 50 if(d[y]==d[x]+1&&e[i].flow) 51 if(dis=makeflow(y,min(flow-maxflow,e[i].flow))){ 52 e[i].flow-=dis; 53 e[e[i].rev].flow+=dis; 54 maxflow+=dis; 55 } 56 } 57 if(!maxflow)d[x]=-1; 58 return maxflow; 59 } 60 int dinic(){ 61 int ans=0,d; 62 while(makelevel()) 63 while(d=makeflow(S,inf)) 64 ans+=d; 65 return ans; 66 } 67 int ID1[330][330],ID2[maxn],a[maxn]; 68 int main(){ 69 freopen(FILE".in","r",stdin); 70 freopen(FILE".out","w",stdout); 71 n=read();m=read(); 72 up(i,1,n)a[i]=read(); 73 up(i,1,n)up(j,i,n){ 74 v[i][j]=read();if(i==j)v[i][j]-=a[i]; 75 ID1[i][j]=++cnt; 76 } 77 up(i,1,n)up(j,i,n)if(i!=j)insert(ID1[i][j],ID1[i][j-1],inf),insert(ID1[i][j],ID1[i+1][j],inf); 78 S=++cnt,T=++cnt; 79 if(m==1)up(i,1,n){ 80 if(!ID2[a[i]]){ 81 ID2[a[i]]=++cnt; 82 insert(ID2[a[i]],T,squ(a[i])); 83 } 84 insert(ID1[i][i],ID2[a[i]],inf); 85 } 86 int Sum=0; 87 up(i,1,n)up(j,i,n){ 88 if(v[i][j]>0)insert(S,ID1[i][j],v[i][j]),Sum+=v[i][j]; 89 else if(v[i][j]<0)insert(ID1[i][j],T,-v[i][j]); 90 } 91 int ans=dinic(); 92 printf("%d\n",Sum-ans); 93 return 0; 94 }
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.
