2019牛客多校第二场
比赛总结：
看错F题意@byf
要勇于打表找规律
题解(不定更新)
A EddyWalker
B EddyWalker2
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define rep(i,a,n) for (int i=a;i<n;i++) 4 #define per(i,a,n) for (int i=n-1;i>=a;i--) 5 #define pb push_back 6 #define mp make_pair 7 #define all(x) (x).begin(),(x).end() 8 #define fi first 9 #define se second 10 #define SZ(x) ((int)(x).size()) 11 typedef vector<int> VI; 12 typedef long long ll; 13 typedef long long LL; 14 typedef pair<int,int> PII; 15 const ll mod=1000000007; 16 ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 17 // head 18 19 ll n; 20 namespace linear_seq { 21 const int N=10010; 22 ll res[N],base[N],_c[N],_md[N]; 23 24 vector<int> Md; 25 void mul(ll *a,ll *b,int k) { 26 rep(i,0,k+k) _c[i]=0; 27 rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod; 28 for (int i=k+k-1;i>=k;i--) if (_c[i]) 29 rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod; 30 rep(i,0,k) a[i]=_c[i]; 31 } 32 int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+... 33 ll ans=0,pnt=0; 34 int k=SZ(a); 35 assert(SZ(a)==SZ(b)); 36 rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1; 37 Md.clear(); 38 rep(i,0,k) if (_md[i]!=0) Md.push_back(i); 39 rep(i,0,k) res[i]=base[i]=0; 40 res[0]=1; 41 while ((1ll<<pnt)<=n) pnt++; 42 for (int p=pnt;p>=0;p--) { 43 mul(res,res,k); 44 if ((n>>p)&1) { 45 for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0; 46 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod; 47 } 48 } 49 rep(i,0,k) ans=(ans+res[i]*b[i])%mod; 50 if (ans<0) ans+=mod; 51 return ans; 52 } 53 VI BM(VI s) { 54 VI C(1,1),B(1,1); 55 int L=0,m=1,b=1; 56 rep(n,0,SZ(s)) { 57 ll d=0; 58 rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod; 59 if (d==0) ++m; 60 else if (2*L<=n) { 61 VI T=C; 62 ll c=mod-d*powmod(b,mod-2)%mod; 63 while (SZ(C)<SZ(B)+m) C.pb(0); 64 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod; 65 L=n+1-L; B=T; b=d; m=1; 66 } else { 67 ll c=mod-d*powmod(b,mod-2)%mod; 68 while (SZ(C)<SZ(B)+m) C.pb(0); 69 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod; 70 ++m; 71 } 72 } 73 return C; 74 } 75 int gao(VI a,ll n) { 76 VI c=BM(a); 77 c.erase(c.begin()); 78 rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod; 79 return solve(n,c,VI(a.begin(),a.begin()+SZ(c))); 80 } 81 }; 82 83 int main() { 84 vector<int>v; 85 int t; 86 scanf("%d",&t); 87 int k;LL n; 88 while(t--) 89 { 90 scanf("%d%lld",&k,&n); 91 if(n==-1) 92 { 93 printf("%lld\n",2LL*powmod(k+1,mod-2)%mod); 94 continue; 95 } 96 v.clear(); 97 ll ook=powmod(k,mod-2); 98 v.push_back(1); 99 for(int i=1;i<=2*k+1;i++) 100 { 101 ll pa=0; 102 for(int j=max(0,i-k);j<i;j++) pa=(pa+1ll*v[j]*ook)%mod; 103 v.push_back(int(pa%mod)); 104 } 105 printf("%lld\n",1LL * linear_seq::gao(v,n) % mod); 106 } 107 }
C Go on Strike!
D Kth Minimum Clique
题解：https://blog.csdn.net/liufengwei1/article/details/97009245
1 #include<bits/stdc++.h> 2 #define maxl 110 3 using namespace std; 4 5 int n,k; 6 long long ans; 7 long long a[maxl]; 8 int mp[maxl][maxl]; 9 bitset <maxl> b[maxl]; 10 char ch[maxl]; 11 struct node 12 { 13 long long w; 14 int id; 15 bitset <maxl> s; 16 bool operator > (const node &y)const 17 { 18 return w>y.w; 19 } 20 }; 21 priority_queue<node,vector<node>,greater<node> >q; 22 23 inline void prework() 24 { 25 scanf("%d%d",&n,&k); 26 for(int i=0;i<n;i++) 27 scanf("%lld",&a[i]); 28 for(int i=0;i<n;i++) 29 { 30 scanf("%s",ch); 31 for(int j=0;j<n;j++) 32 { 33 mp[i][j]=ch[j]-'0'; 34 b[i][j]=mp[i][j]; 35 } 36 } 37 } 38 39 inline void mainwork() 40 { 41 node d;bitset <maxl> ds; 42 q.push(node{0,0,ds}); 43 44 while(k>0 && !q.empty()) 45 { 46 d=q.top();q.pop(); 47 k--; 48 if(!k) 49 { 50 ans=d.w; 51 return; 52 } 53 for(int i=d.id;i<n;i++) 54 if(!d.s[i] && (d.s&b[i])==d.s) 55 { 56 d.s[i]=1; 57 q.push(node{a[i]+d.w,i,d.s}); 58 d.s[i]=0; 59 } 60 } 61 ans=-1; 62 } 63 64 inline void print() 65 { 66 printf("%lld",ans); 67 } 68 69 int main() 70 { 71 prework(); 72 mainwork(); 73 print(); 74 return 0; 75 }
E MAZE
F Partition problem
题解：https://blog.csdn.net/liufengwei1/article/details/96729137
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int n; 5 long long ans; 6 int aa[30],bb[30]; 7 int v[30][30]; 8 long long f[30][1<<18]; 9 long long sum[30]; 10 int num[1<<18]; 11 12 inline void prework() 13 { 14 for(int i=0,x=1;i<18;i++,x<<=1) 15 num[x]=i; 16 scanf("%d",&n); 17 for(int i=0;i<2*n;i++) 18 for(int j=0;j<2*n;j++) 19 scanf("%d",&v[i][j]),sum[i+1]+=v[i][j]; 20 for(int i=1;i<=2*n;i++) 21 sum[i]=sum[i-1]+sum[i]; 22 int l,s,d; 23 for(int i=0;i<2*n;i++) 24 { 25 l=1<<(min(2*n,18)); 26 for(int j=1;j<l;j++) 27 { 28 s=j; 29 while(s) 30 { 31 d=s&-s; 32 f[i][j]+=v[i][num[d]]; 33 s-=d; 34 } 35 } 36 } 37 } 38 39 inline void dfs(int k,int as,int bs,int acnt,long long w) 40 { 41 if(acnt+2*n-1-k+1<n || k-acnt+2*n-1-k+1<n) 42 return; 43 if(w+sum[2*n]-sum[k+1-1]<ans) 44 return; 45 if(w>ans) 46 ans=w; 47 if(k>=2*n) 48 return; 49 long long tmp=0;int tas,tbs; 50 if(k>=18) 51 { 52 aa[++aa[0]]=k; 53 tmp=w+f[k][bs]; 54 for(int i=1;i<=bb[0];i++) 55 tmp+=v[k][bb[i]]; 56 dfs(k+1,as,bs,acnt+1,tmp); 57 aa[aa[0]--]=0; 58 59 bb[++bb[0]]=k; 60 tmp=w+f[k][as]; 61 for(int i=1;i<=aa[0];i++) 62 tmp+=v[k][aa[i]]; 63 dfs(k+1,as,bs,acnt,tmp); 64 bb[bb[0]--]=0; 65 } 66 else 67 { 68 dfs(k+1,as|(1<<k),bs,acnt+1,w+f[k][bs]); 69 dfs(k+1,as,bs|(1<<k),acnt,w+f[k][as]); 70 } 71 } 72 73 inline void mainwork() 74 { 75 dfs(0,0,0,0,0); 76 } 77 78 inline void print() 79 { 80 printf("%lld",ans); 81 } 82 83 int main() 84 { 85 prework(); 86 mainwork(); 87 print(); 88 return 0; 89 }
G Polygons
H Second Large Rectangle
题解：https://blog.csdn.net/liufengwei1/article/details/96730426
1 #include<bits/stdc++.h> 2 #define maxl 1010 3 4 using namespace std; 5 6 int n,m,top,mx,mx1,my1,mx2,my2,secmx; 7 int s[maxl],h[maxl],l[maxl],r[maxl]; 8 int a[maxl][maxl]; 9 char ch[maxl]; 10 11 inline void prework() 12 { 13 scanf("%d%d",&n,&m); 14 for(int i=1;i<=n;i++) 15 { 16 scanf("%s",ch+1); 17 for(int j=1;j<=m;j++) 18 a[i][j]=ch[j]-'0'; 19 } 20 } 21 22 inline void solve1() 23 { 24 mx=0;int tmp; 25 for(int i=1;i<=m;i++) 26 h[i]=0; 27 for(int i=1;i<=n;i++) 28 { 29 for(int j=1;j<=m;j++) 30 if(a[i][j]==1) 31 h[j]++; 32 else 33 h[j]=0; 34 top=0; 35 for(int j=1;j<=m;j++) 36 { 37 while(top>0 && h[s[top]]>h[j]) 38 r[s[top]]=j-1,--top; 39 l[j]=s[top]+1; 40 s[++top]=j; 41 } 42 while(top>0) 43 r[s[top]]=m,--top; 44 for(int j=1;j<=m;j++) 45 if(h[j]) 46 { 47 tmp=h[j]*(r[j]-l[j]+1); 48 if(tmp>mx) 49 { 50 mx=tmp; 51 mx1=i-h[j]+1;my1=l[j]; 52 mx2=i;my2=r[j]; 53 } 54 } 55 } 56 } 57 58 inline void solve2() 59 { 60 secmx=0;int tmp,tx1,tx2,ty1,ty2; 61 for(int i=1;i<=m;i++) 62 h[i]=0; 63 for(int i=1;i<=n;i++) 64 { 65 for(int j=1;j<=m;j++) 66 if(a[i][j]==1) 67 h[j]++; 68 else 69 h[j]=0; 70 top=0; 71 for(int j=1;j<=m;j++) 72 { 73 while(top>0 && h[s[top]]>h[j]) 74 r[s[top]]=j-1,--top; 75 l[j]=s[top]+1; 76 s[++top]=j; 77 } 78 while(top>0) 79 r[s[top]]=m,--top; 80 for(int j=1;j<=m;j++) 81 if(h[j]) 82 { 83 tx1=i-h[j]+1;ty1=l[j]; 84 tx2=i;ty2=r[j]; 85 tmp=h[j]*(r[j]-l[j]+1); 86 if(mx1!=tx1 || my1!=ty1 || mx2!=tx2 || my2!=ty2) 87 secmx=max(tmp,secmx); 88 secmx=max(secmx,(h[j]-1)*(r[j]-l[j]+1)); 89 secmx=max(secmx,h[j]*(r[j]-l[j])); 90 } 91 } 92 } 93 94 inline void mainwork() 95 { 96 solve1(); 97 solve2(); 98 } 99 100 inline void print() 101 { 102 printf("%d\n",secmx); 103 } 104 105 int main() 106 { 107 prework(); 108 mainwork(); 109 print(); 110 return 0; 111 }
I Inside A Rectangle
J Subarray
题解：https://blog.csdn.net/liufengwei1/article/details/96836978
1 //牛逼网友的J 2 #include<bits/stdc++.h> 3 using namespace std; 4 typedef long long ll; 5 const int INF = 0x3f3f3f3f; 6 const int MAXN = 10000000, MAXM = 1000000; 7 8 int l[MAXM + 5], r[MAXM + 5], f[MAXM + 5], g[MAXM + 5]; 9 int sum[MAXN * 3 + 5], b[MAXN * 3 + 5], c[MAXN * 3 + 5]; 10 11 int main() 12 { 13 int n; 14 scanf("%d",&n); 15 for(int i=1;i<=n;i++) 16 scanf("%d%d",&l[i],&r[i]); 17 f[1]=r[1]-l[1]+1; 18 //f[i]以i段右端点为结尾的能构造出的最大的前缀和 19 for(int i=2;i<=n;i++) 20 f[i]=max(0,f[i-1]-(l[i]-r[i-1]-1))+r[i]-l[i]+1; 21 //0：以i-1段右端点结尾的能构造出的最大的前缀和都不足够跨过[i-1,i]之间的-1 22 //f[i - 1] - (l[i] - r[i - 1] - 1)：跨过之后还剩下多少贡献给这段 23 g[n]=r[n]-l[n]+1; 24 //g[i]以i段左端点为开头的能构造出的最大的前缀和 25 for(int i=n-1;i>=1;i--) 26 g[i]=max(0,g[i+1]-(l[i+1]-r[i]-1))+r[i]-l[i]+1; 27 //ERR1(f, n); 28 //ERR1(g, n); 29 int i=1,base=10000000; 30 ll ans=0; 31 while(i<=n) 32 { 33 int j=i+1; 34 while(j<=n && g[j]+f[j-1]>=l[j]-r[j - 1]-1) { 35 //说明这个[j-1,j]之间的-1段可以因为两侧的f[j-1]和g[j]足够大而连接起来 36 j++; 37 } 38 j--; 39 //此时j是从i开始最远能够连接到的区间 40 int left=max(0,l[i]-g[i]),right=min(1000000000-1,r[j]+f[j]); 41 //left,right是至少会产生一个贡献的范围 42 //ERR(left, right); 43 int t=i,mi=INF,mx=0; 44 sum[0]=0; 45 for(int k=left;k<=right;k++) 46 { 47 //统计这一整段可连接区间的前缀和 48 if(k>=l[t] && k<=r[t]) 49 sum[k-left+1]=sum[k-left]+1; 50 else 51 sum[k-left+1]=sum[k-left]-1; 52 if(k==r[t]) 53 t++; 54 mi=min(mi,sum[k-left+1]+base); 55 mx=max(mx,sum[k-left+1]+base); 56 //b记录前缀和出现过的次数 57 b[sum[k-left+1]+base]++; 58 } 59 //ERR1(sum, right); 60 //b记录前缀和出现过的次数的后缀和 61 for(int k=mx-1;k>=mi;k--) 62 b[k]+=b[k+1]; 63 //包含最左侧点的贡献 64 ans+=b[base+1]; 65 for(int k=left;k<=right;k++) { 66 t=sum[k-left+1]+base; 67 //t表示k位置sum的值 68 //b[t+1]比t大的值的个数 69 //c[t+1]比在k位置左侧的比t大的值的个数的lazy 70 b[t+1]-=c[t+1]; //把lazy加上去 71 c[t]+=c[t+1]+1; //lazy标记下移 72 c[t+1] = 0; //清空lazy 73 ans+=b[t+1]; 74 } 75 for(int k=mi;k<=mx;k++) 76 b[k]=0,c[k]=0; 77 i=j+1; 78 } 79 printf("%lld", ans); 80 return 0; 81 }
