2018 Multi-University Training Contest 1 hdu搬运工
2018-07-26 00:26:29http://acm.hdu.edu.cn/contests/contest_show.php?cid=802
被三整除一定是三个数相等，其次是被2整除且被四整除的数可取
s=2x1;
s=4x2;
s=x1+2x2
A题
// A #include <bits/stdc++.h> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int i=n-1;i>=a;i--) #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second #define SZ(x) ((int)(x).size()) typedef vector<int> VI; typedef long long ll; typedef pair<int,int> PII; const ll mod=1000000007; 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;} ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} // head int n,_; int main() { for (scanf("%d",&_);_;_--) { scanf("%d",&n); if (n%3==0) printf("%lld\n",1ll*n*n*n/27); else if (n%4==0) printf("%lld\n",1ll*n*n*n/32); else puts("-1"); }
B 题
1 // B 2 #include <stdio.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <algorithm> 7 using namespace std; 8 const int MAXN = 101010; 9 const int MAXS = 101010; 10 int n; 11 class Str 12 { 13 public: 14 int l, r, add; 15 bool operator <(const Str &b) const 16 { 17 if(l >= r && b.l < b.r) 18 return false; 19 if(l < r && b.l >= b.r) 20 return true; 21 if(l >= r && b.l >= b.r) 22 return r > b.r; 23 return l < b.l; 24 } 25 }a[MAXN]; 26 char s[MAXS]; 27 void solve() 28 { 29 scanf("%d", &n); 30 for(int i = 1; i <= n; i++) 31 { 32 scanf("%s", s); 33 int len = strlen(s); 34 a[i].l = a[i].r = a[i].add = 0; 35 for(int j = 0; j < len; j++) 36 { 37 if(s[j] == '(') 38 a[i].r++; 39 else 40 { 41 if(a[i].r > 0) 42 a[i].r--, a[i].add++; 43 else 44 a[i].l++; 45 } 46 } 47 } 48 sort(a + 1, a + n + 1); 49 int ans = 0; 50 int now = 0; 51 for(int i = 1; i <= n; i++) 52 { 53 if(a[i].l > now) 54 a[i].l = now; 55 ans += a[i].l + a[i].add; 56 now -= a[i].l; 57 now += a[i].r; 58 } 59 printf("%d\n", ans * 2); 60 } 61 int main() 62 { 63 int T; 64 scanf("%d", &T); 65 for(int t1 = 1; t1 <= T; t1++) 66 solve(); 67 return 0; 68 }
C题
1 // C 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i<n;i++) 5 #define per(i,a,n) for (int i=n-1;i>=a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first 10 #define se second 11 #define SZ(x) ((int)(x).size()) 12 typedef vector<int> VI; 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 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 18 // head 19 20 const int N=101000; 21 int _,n,x,y; 22 pair<PII,int> p[N]; 23 int main() { 24 for (scanf("%d",&_);_;_--) { 25 scanf("%d",&n); 26 rep(i,0,3*n) { 27 scanf("%d%d",&x,&y); 28 p[i]=mp(mp(x,y),i); 29 } 30 sort(p,p+3*n); 31 rep(i,0,n) { 32 printf("%d %d %d\n",p[3*i].se+1,p[3*i+1].se+1,p[3*i+2].se+1); 33 } 34 } 35 }
D题
1 // D 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i<n;i++) 5 #define per(i,a,n) for (int i=n-1;i>=a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first 10 #define se second 11 #define SZ(x) ((int)(x).size()) 12 typedef vector<int> VI; 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 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 18 // head 19 20 const int N=101000; 21 int _,n,m,pre[N],l,r,ret[N]; 22 int main() { 23 for (scanf("%d",&_);_;_--) { 24 scanf("%d%d",&n,&m); 25 rep(i,1,n+1) pre[i]=i; 26 rep(i,0,m) { 27 scanf("%d%d",&l,&r); 28 pre[r]=min(pre[r],l); 29 } 30 per(i,1,n) pre[i]=min(pre[i],pre[i+1]); 31 int pl=1; 32 set<int> val; 33 rep(i,1,n+1) val.insert(i); 34 rep(i,1,n+1) { 35 while (pl<pre[i]) { 36 val.insert(ret[pl]); 37 pl++; 38 } 39 ret[i]=*val.begin(); 40 val.erase(ret[i]); 41 } 42 rep(i,1,n+1) printf("%d%c",ret[i]," \n"[i==n]); 43 } 44 }
E题
1 // E 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i<n;i++) 5 #define per(i,a,n) for (int i=n-1;i>=a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first 10 #define se second 11 #define SZ(x) ((int)(x).size()) 12 typedef vector<int> VI; 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 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 18 // head 19 20 const int N=201000; 21 struct tSPnode { 22 int ty,u,v,w; 23 ll dp[2][2]; 24 ll way[2][2]; 25 tSPnode *l,*r; 26 }tpool[N*2],*tcur,*E[N]; 27 int dq[N],inq[N],_,n,m,u,v,w; 28 map<int,tSPnode*> se[N]; 29 tSPnode* newtnode(int u,int v,int w=0) { 30 tSPnode* p=tcur++; 31 p->ty=-2; p->u=u; p->v=v; p->l=p->r=0; p->w=w; 32 memset(p->dp,0xee,sizeof(p->dp)); 33 memset(p->way,0,sizeof(p->way)); 34 return p; 35 } 36 void insertSP(int u,int v,tSPnode *c) { 37 if (se[u].count(v)) { 38 tSPnode *p=newtnode(u,v); 39 p->ty=-1; // par 40 p->l=se[u][v]; 41 p->r=c; 42 se[u][v]=se[v][u]=p; 43 } else { 44 se[u][v]=se[v][u]=c; 45 } 46 } 47 48 void gao(tSPnode *Rt) { 49 if (Rt->ty==-2) { 50 Rt->dp[0][0]=0; 51 Rt->way[0][0]=1; 52 Rt->dp[1][1]=Rt->w; 53 Rt->way[1][1]=1; 54 } else if (Rt->ty==-1) { 55 if (Rt->l->u!=Rt->u) swap(Rt->l->u,Rt->l->v); 56 if (Rt->r->v!=Rt->v) swap(Rt->r->u,Rt->r->v); 57 assert(Rt->l->u==Rt->u); 58 assert(Rt->r->u==Rt->u); 59 assert(Rt->l->v==Rt->v); 60 assert(Rt->r->v==Rt->v); 61 gao(Rt->l); gao(Rt->r); 62 rep(pl,0,3) rep(pr,0,3) { 63 int fl=(pl==1),gl=(pl==2),fr=(pr==1),gr=(pr==2); 64 if (Rt->dp[fl+gl][fr+gr]<Rt->l->dp[fl][fr]+Rt->r->dp[gl][gr]) { 65 Rt->dp[fl+gl][fr+gr]=Rt->l->dp[fl][fr]+Rt->r->dp[gl][gr]; 66 Rt->way[fl+gl][fr+gr]=0; 67 } 68 if (Rt->dp[fl+gl][fr+gr]==Rt->l->dp[fl][fr]+Rt->r->dp[gl][gr]) { 69 (Rt->way[fl+gl][fr+gr]+=Rt->l->way[fl][fr]*Rt->r->way[gl][gr])%=mod; 70 } 71 } 72 } else { 73 if (Rt->l->u!=Rt->u&&Rt->l->v!=Rt->u) swap(Rt->l,Rt->r); 74 if (Rt->l->u!=Rt->u) swap(Rt->l->u,Rt->l->v); 75 if (Rt->r->v!=Rt->v) swap(Rt->r->u,Rt->r->v); 76 // printf("%d %d %d %d %d %d\n",Rt->u,Rt->v,Rt->l->u,Rt->l->v,Rt->r->u,Rt->r->v); 77 assert(Rt->l->v==Rt->r->u); 78 gao(Rt->l); gao(Rt->r); 79 rep(fl,0,2) rep(gr,0,2) rep(fmid,0,3) { 80 int gl=(fmid==1),fr=(fmid==2); 81 if (Rt->dp[fl][gr]<Rt->l->dp[fl][fr]+Rt->r->dp[gl][gr]) { 82 Rt->dp[fl][gr]=Rt->l->dp[fl][fr]+Rt->r->dp[gl][gr]; 83 Rt->way[fl][gr]=0; 84 } 85 if (Rt->dp[fl][gr]==Rt->l->dp[fl][fr]+Rt->r->dp[gl][gr]) { 86 (Rt->way[fl][gr]+=Rt->l->way[fl][fr]*Rt->r->way[gl][gr])%=mod; 87 } 88 } 89 } 90 // printf("%d %d %d\n",Rt->u,Rt->v,Rt->ty); 91 // printf("%lld %lld %lld %lld\n",Rt->dp[0][0],Rt->dp[0][1],Rt->dp[1][0],Rt->dp[1][1]); 92 } 93 void parse() { 94 rep(i,0,n) se[i].clear(); 95 rep(i,0,m) { 96 int u=E[i]->u,v=E[i]->v; 97 insertSP(u,v,E[i]); 98 } 99 int tot=0; 100 rep(u,0,n) { 101 inq[u]=0; 102 if (SZ(se[u])==2) dq[tot++]=u,inq[u]=1; 103 } 104 VI rv; 105 rep(i,0,tot) { 106 int u=dq[i]; 107 if (SZ(se[u])!=2) { rv.pb(u); continue;} 108 auto it=se[u].begin(),nit=it; 109 ++it; 110 int p=it->fi,q=nit->fi; 111 tSPnode *s=it->se,*t=nit->se; 112 se[p].erase(u); se[q].erase(u); 113 tSPnode *r=newtnode(p,q); r->ty=u; 114 r->l=s, r->r=t; 115 insertSP(p,q,r); 116 if (!inq[p]&&SZ(se[p])==2) dq[tot++]=p,inq[p]=1; 117 if (!inq[q]&&SZ(se[q])==2) dq[tot++]=q,inq[q]=1; 118 } 119 rep(i,0,n) if (!inq[i]) rv.pb(i); 120 assert(SZ(rv)==2); 121 tSPnode *Rt=se[rv[0]][rv[1]]; 122 gao(Rt); 123 ll ret=-1,way=0; 124 rep(i,0,2) rep(j,0,2) { 125 if (ret<Rt->dp[i][j]) { 126 ret=Rt->dp[i][j]; 127 way=0; 128 } 129 if (Rt->dp[i][j]==ret) way=(way+Rt->way[i][j])%mod; 130 } 131 printf("%lld %lld\n",ret,way); 132 } 133 134 void gao() { 135 scanf("%d%d",&n,&m); 136 tcur=tpool; 137 rep(i,0,m) { 138 scanf("%d%d%d",&u,&v,&w); 139 --u; --v; 140 E[i]=newtnode(u,v,w); 141 } 142 parse(); 143 } 144 145 int main() { 146 for (scanf("%d",&_);_;_--) gao(); 147 } 148
F题
1 // F 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i<n;i++) 5 #define per(i,a,n) for (int i=n-1;i>=a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first 10 #define se second 11 #define SZ(x) ((int)(x).size()) 12 typedef vector<int> VI; 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 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 18 // head 19 20 namespace polysum { 21 const int D=101000; 22 ll a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][2],C[D]; 23 ll calcn(int d,ll *a,ll n) { 24 if (n<=d) return a[n]; 25 p1[0]=p2[0]=1; 26 rep(i,0,d+1) { 27 ll t=(n-i+mod)%mod; 28 p1[i+1]=p1[i]*t%mod; 29 } 30 rep(i,0,d+1) { 31 ll t=(n-d+i+mod)%mod; 32 p2[i+1]=p2[i]*t%mod; 33 } 34 ll ans=0; 35 rep(i,0,d+1) { 36 ll t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod; 37 if ((d-i)&1) ans=(ans-t+mod)%mod; 38 else ans=(ans+t)%mod; 39 } 40 return ans; 41 } 42 void init(int M) { 43 f[0]=f[1]=g[0]=g[1]=1; 44 rep(i,2,M+5) f[i]=f[i-1]*i%mod; 45 g[M+4]=powmod(f[M+4],mod-2); 46 per(i,1,M+4) g[i]=g[i+1]*(i+1)%mod; 47 } 48 ll polysum(ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i] 49 a[m+1]=calcn(m,a,m+1); 50 rep(i,1,m+2) a[i]=(a[i-1]+a[i])%mod; 51 return calcn(m+1,a,n-1); 52 } 53 ll qpolysum(ll R,ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i 54 if (R==1) return polysum(n,a,m); 55 a[m+1]=calcn(m,a,m+1); 56 ll r=powmod(R,mod-2),p3=0,p4=0,c,ans; 57 h[0][0]=0;h[0][1]=1; 58 rep(i,1,m+2) { 59 h[i][0]=(h[i-1][0]+a[i-1])*r%mod; 60 h[i][1]=h[i-1][1]*r%mod; 61 } 62 rep(i,0,m+2) { 63 ll t=g[i]*g[m+1-i]%mod; 64 if (i&1) p3=((p3-h[i][0]*t)%mod+mod)%mod,p4=((p4-h[i][1]*t)%mod+mod)%mod; 65 else p3=(p3+h[i][0]*t)%mod,p4=(p4+h[i][1]*t)%mod; 66 } 67 c=powmod(p4,mod-2)*(mod-p3)%mod; 68 rep(i,0,m+2) h[i][0]=(h[i][0]+h[i][1]*c)%mod; 69 rep(i,0,m+2) C[i]=h[i][0]; 70 ans=(calcn(m,C,n)*powmod(R,n)-c)%mod; 71 if (ans<0) ans+=mod; 72 return ans; 73 } 74 } 75 76 ll mul(ll a,ll b,ll c) { 77 a%=mod; b%=mod; c%=mod; 78 return a*b%mod*c%mod; 79 } 80 81 const int N=20100; 82 int n,x,pos[N],_; 83 pair<int,int> p[N]; 84 ll a,b,fl[N],fr[N],s[N]; 85 void solve() { 86 scanf("%d%lld%lld",&n,&a,&b); 87 rep(i,0,n) { 88 scanf("%d",&x); 89 p[i]=mp(x,i); 90 } 91 ll ret=0; 92 sort(p,p+n); 93 rep(rem,0,n) { 94 int t=0; 95 rep(i,0,n) if (p[i].fi%n==rem) pos[t++]=p[i].se-(p[i].fi-rem); 96 sort(pos,pos+t); 97 rep(i,0,t) { 98 fl[i]=(a-pos[i]+n-1)/n; 99 fr[i]=(b-pos[i])/n; 100 // printf("%d %lld %lld\n",i,fl[i],fr[i]); 101 } 102 rep(i,0,t) if (fl[i]<=fr[i]) { 103 rep(pl,0,5) { 104 ll x=fl[i]+pl; 105 s[pl]=mul(rem+x*n,(pos[i]+x*n)-a+1,b-(pos[i]+x*n)+1); 106 if (pl>0) s[pl]=(s[pl]+s[pl-1])%mod; 107 if (s[pl]<0) s[pl]+=mod; 108 } 109 ret=(ret+polysum::calcn(4,s,fr[i]-fl[i]))%mod; 110 // printf("%d %d %lld\n",i,i,polysum::calcn(4,s,fr[i]-fl[i])); 111 } 112 rep(i,0,t) rep(j,i+1,t) if (max(fl[i],fl[j])<=min(fr[i],fr[j])) { 113 ll l=max(fl[i],fl[j]),r=min(fr[i],fr[j]); 114 rep(pl,0,5) { 115 ll x=l+pl; 116 s[pl]=mul(2*(rem+x*n),(pos[i]+x*n)-a+1,b-(pos[j]+x*n)+1); 117 if (pl>0) s[pl]=(s[pl]+s[pl-1])%mod; 118 if (s[pl]<0) s[pl]+=mod; 119 } 120 ret=(ret+polysum::calcn(4,s,r-l))%mod; 121 // printf("%d %d %lld\n",i,j,polysum::calcn(4,s,r-l)); 122 } 123 } 124 printf("%lld\n",ret); 125 } 126 int main() { 127 polysum::init(10); 128 for (scanf("%d",&_);_;_--) solve(); 129 }
G 题
1 // G 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i<n;i++) 5 #define per(i,a,n) for (int i=n-1;i>=a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first 10 #define se second 11 #define SZ(x) ((int)(x).size()) 12 typedef vector<int> VI; 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 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 18 // head 19 20 //int a[101000]; 21 int _; 22 ll n; 23 ll mul(ll a,ll b) { 24 a%=mod; b%=mod; 25 // printf("%lld %lld %lld %lld\n",a,b,c,d); 26 return a*b%mod; 27 } 28 29 30 ll mul(ll a,ll b,ll c) { 31 a%=mod; b%=mod; c%=mod; 32 // printf("%lld %lld %lld %lld\n",a,b,c,d); 33 return a*b%mod*c%mod; 34 } 35 int main() { 36 for (scanf("%d",&_);_;_--) { 37 scanf("%lld",&n); 38 ll l=0,r=n; 39 while (l+1<r) { 40 ll md=(l+r)>>1; 41 if (2*md-__builtin_popcountll(md)<n) l=md; else r=md; 42 } 43 ll ans=(1+mul(n-1+n+1-2*r,r,(mod+1)/2))%mod; 44 rep(i,0,61) { 45 ll q=r/(1ll<<(i+1)),q2=r-(q<<(i+1)); 46 if (q2<(1ll<<i)) q2=0; else q2-=(1ll<<i); 47 ans=(ans+mul(q,1ll<<i)+q2)%mod; 48 } 49 /* rep(k,0,r) { 50 ans+=__builtin_popcount(k); 51 }*/ 52 printf("%lld\n",ans); 53 } 54 }
H题
1 // H 2 3 #include <bits/stdc++.h> 4 using namespace std; 5 #define rep(i,a,n) for (int i=a;i<n;i++) 6 #define per(i,a,n) for (int i=n-1;i>=a;i--) 7 #define pb push_back 8 #define mp make_pair 9 #define all(x) (x).begin(),(x).end() 10 #define fi first 11 #define se second 12 #define SZ(x) ((int)(x).size()) 13 typedef vector<int> VI; 14 typedef long long ll; 15 typedef pair<int,int> PII; 16 const ll mod=1000000007; 17 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;} 18 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 19 // head 20 21 const int N=1010000; 22 int stk[N],top,l[N],r[N],vis[N],n,x,_; 23 PII a[N]; 24 ll inv[N],ret; 25 26 27 int dfs(int u) { 28 int s=1; 29 if (l[u]) s+=dfs(l[u]); 30 if (r[u]) s+=dfs(r[u]); 31 ret=ret*inv[s]%mod; 32 return s; 33 } 34 void build() { 35 int top=0; 36 rep(i,1,n+1) l[i]=0,r[i]=0,vis[i]=0; 37 rep(i,1,n+1) { 38 int k=top; 39 while (k>0&&a[stk[k-1]]>a[i]) --k; 40 if (k) r[stk[k-1]]=i; 41 if (k<top) l[i]=stk[k]; 42 stk[k++]=i; 43 top=k; 44 } 45 rep(i,1,n+1) vis[l[i]]=vis[r[i]]=1; 46 int rt=0; 47 rep(i,1,n+1) if (vis[i]==0) rt=i; 48 dfs(rt); 49 } 50 51 int main() { 52 inv[1]=1; 53 rep(i,2,1000001) inv[i]=inv[mod%i]*(mod-mod/i)%mod; 54 for (scanf("%d",&_);_;_--) { 55 scanf("%d",&n); 56 rep(i,1,n+1) { 57 scanf("%d",&x); 58 a[i]=mp(-x,i); 59 } 60 ret=inv[2]*n%mod; 61 build(); 62 printf("%lld\n",ret); 63 } 64 }
I题
1 // I 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i<n;i++) 5 #define per(i,a,n) for (int i=n-1;i>=a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first 10 #define se second 11 #define SZ(x) ((int)(x).size()) 12 typedef vector<int> VI; 13 typedef long long ll; 14 typedef pair<int,int> PII; 15 const ll mod1=1000000007; 16 const ll mod2=1000000009; 17 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 18 // head 19 20 const int L=601000,N=101000;; 21 char s[L]; 22 int ss[L],sa[L],rk[L],bp[N],mg[N]; 23 24 struct substr { 25 int id,l,len; 26 }; 27 typedef pair<int,int> hashv; 28 vector<substr> g[N],sg[N]; 29 hashv base(13331,23333); 30 hashv hv[L],pw[L]; 31 32 int n,m,u,v,_; 33 34 hashv operator + (hashv a,hashv b) { 35 int c1=a.fi+b.fi,c2=a.se+b.se; 36 if (c1>=mod1) c1-=mod1; 37 if (c2>=mod2) c2-=mod2; 38 return mp(c1,c2); 39 } 40 41 hashv operator - (hashv a,hashv b) { 42 int c1=a.fi-b.fi,c2=a.se-b.se; 43 if (c1<0) c1+=mod1; 44 if (c2<0) c2+=mod2; 45 return mp(c1,c2); 46 } 47 48 hashv operator * (hashv a,hashv b) { 49 return mp(1ll*a.fi*b.fi%mod1,1ll*a.se*b.se%mod2); 50 } 51 52 vector<substr> lyndon(char *s,int n,int id,int &mg){ 53 int k=0; 54 mg=0; 55 vector<substr> g; 56 while (k<n) { 57 int i=k+1,j=k+2; 58 while (1) { 59 if (j==n+1||s[j-1]<s[i-1]) { 60 int mul=0,b=k,l=j-i; 61 while (k<i) { 62 mul++; 63 k+=j-i; 64 } 65 g.pb((substr){id,b,l}); 66 mg=max(mg,l); 67 break; 68 } else { 69 if (s[j-1]>s[i-1]) { 70 i=k+1; 71 } else { 72 i=i+1; 73 } 74 j=j+1; 75 } 76 } 77 } 78 return g; 79 } 80 81 void buildSA(int *s,int *sa,int *rk,int n,int m=128) { 82 static int X[L],Y[L],c[L]; 83 int *x=X,*y=Y; 84 rep(i,0,m) c[i]=0; 85 rep(i,0,n) c[x[i]=s[i]]++; 86 rep(i,1,m) c[i]+=c[i-1]; 87 per(i,0,n) sa[--c[x[i]]]=i; 88 for (int k=1;k<n;k<<=1) { 89 int p=0; 90 per(i,n-k,n) y[p++]=i; 91 rep(i,0,n) if (sa[i]>=k) y[p++]=sa[i]-k; 92 rep(i,0,m) c[i]=0; 93 rep(i,0,n) c[x[y[i]]]++; 94 rep(i,1,m) c[i]+=c[i-1]; 95 per(i,0,n) sa[--c[x[y[i]]]]=y[i]; 96 swap(x,y); 97 p=1; x[sa[0]]=0; y[n]=-1; 98 rep(i,1,n) x[sa[i]]=y[sa[i-1]]==y[sa[i]]&& 99 y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++; 100 if (p==n) break; 101 m=p; 102 } 103 rep(i,0,n)rk[sa[i]]=i; 104 rep(i,0,n) hv[i+1]=hv[i]*base+mp(s[i],s[i]); 105 } 106 107 int pos(const substr &a) { 108 return bp[a.id]+a.l; 109 } 110 111 hashv query(const substr &a) { 112 int posl=bp[a.id]+a.l,posr=bp[a.id]+a.l+a.len; 113 return hv[posr]-hv[posl]*pw[posr-posl]; 114 } 115 int cmp(const substr& a,const substr& b) { 116 int l=min(a.len,b.len); 117 auto p1=query((substr){a.id,a.l,l}),p2=query((substr){b.id,b.l,l}); 118 if (p1==p2) { 119 if (a.len<b.len) return -1; 120 else if (a.len==b.len) return 0; 121 else return 1; 122 } 123 return rk[pos(a)]<rk[pos(b)]?-1:1; 124 } 125 126 substr tmpa[20],tmpb[20]; 127 int tmp[20]; 128 int cmp(substr* a,int p1,substr* b,int p2) { 129 int q1=0,q2=0; 130 while (q1<p1&&q2<p2) { 131 if (a[q1].len==b[q2].len) { 132 int r=cmp(a[q1],b[q2]); 133 if (r==0) q1++,q2++; 134 else return r; 135 } else if (a[q1].len<b[q2].len) { 136 substr c=b[q2]; 137 c.len=a[q1].len; 138 int r=cmp(a[q1],c); 139 if (r!=0) return r; 140 else { 141 b[q2].l+=a[q1].len; 142 b[q2].len-=a[q1].len; 143 q1++; 144 } 145 } else { 146 substr c=a[q1]; 147 c.len=b[q2].len; 148 int r=cmp(c,b[q2]); 149 if (r!=0) return r; 150 else { 151 a[q1].l+=b[q2].len; 152 a[q1].len-=b[q2].len; 153 q2++; 154 } 155 } 156 } 157 if (q1==p1&&q2==p2) return 0; 158 if (q1==p1) return -1; 159 else return 1; 160 } 161 162 substr t1[10],t2[10]; 163 void solve() { 164 scanf("%d%d",&n,&m); 165 int cur=0; 166 rep(i,0,n) { 167 scanf("%s",s); 168 bp[i]=cur; 169 int len=strlen(s); 170 rep(j,0,len) ss[cur++]=s[j]; 171 ss[cur++]=129+i; 172 g[i]=lyndon(s,len,i,mg[i]); 173 sg[i].clear(); 174 for (auto p:g[i]) sg[i].pb((substr){i,p.l,len-p.l}); 175 } 176 buildSA(ss,sa,rk,cur,130+n); 177 rep(i,0,m) { 178 scanf("%d%d",&u,&v); 179 --u; --v; 180 int r=cmp(g[u].back(),g[v][0]); 181 if (r>=0) { 182 printf("%d\n",max(mg[u],mg[v])); 183 } else { 184 int l=-1,r=SZ(g[u])-1; 185 substr sr=sg[v][0]; 186 while (l+1<r) { 187 int md=(l+r)>>1; 188 t1[0]=sg[u][md]; t1[1]=sr; 189 t2[0]=sg[u][md+1]; t2[1]=sr; 190 if (cmp(t1,2,t2,2)==-1) r=md; else l=md; 191 } 192 int pl=0,pr=SZ(g[v]); 193 while (pl+1<pr) { 194 int md=(pl+pr)>>1; 195 t1[0]=sg[v][md]; 196 t2[0]=sg[u][r]; t2[1]=sr; 197 if (cmp(t1,1,t2,2)==1) pl=md; else pr=md; 198 } 199 int ans=sg[u][r].len; 200 if (pl==SZ(sg[v])-1) ans+=sg[v][0].len; 201 else ans+=sg[v][pl+1].l; 202 printf("%d\n",max(ans,max(mg[u],mg[v]))); 203 } 204 } 205 } 206 int main() { 207 pw[0]=mp(1,1); 208 rep(i,1,600001) pw[i]=pw[i-1]*base; 209 for (scanf("%d",&_);_;_--) solve(); 210 }
J题
// J #include <bits/stdc++.h> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int i=n-1;i>=a;i--) #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second #define SZ(x) ((int)(x).size()) typedef vector<int> VI; typedef long long ll; typedef pair<int,int> PII; const ll mod=1000000007; 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;} ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} // head const int N=1010000; int n,_; bool f[N],g[N],preg[N],preg2[N]; char s[N]; int ret[N],ret2[N],cnt[N],cnt2[N]; void gao(bool *g,int *ret,int n) { rep(i,1,n+1) ret[i]=1<<30; int fo=n+1,lo=-1; rep(i,1,n+1) { if (g[i]) { if (fo==n+1) fo=i; lo=i; } } if (fo>lo) { rep(i,1,n+1) ret[i]=0; return; } rep(i,1,n+1) { preg[i]=preg[i-1]^g[i]^1; preg2[i]=preg2[i-1]^g[i]; cnt[i]=cnt[i-1]+preg[i]; cnt2[i]=cnt2[i-1]+preg2[i]; } rep(i,0,lo+1) { if (i<=fo) { int fw=i,lw=lo; if (i==lo) { ret[i]=min(ret[i],3); continue; } int ans=lw-fw; if (preg[fw-1]) ans+=2*(cnt[lw-1]-cnt[fw-1]); else ans+=2*(lw-fw-cnt[lw-1]+cnt[fw-1]); if (preg[lw-1]^preg[fw-1]^1) ans--; ret[i]=min(ret[i],ans); } else { int fw=fo,lw=lo; int ans=i-fo+lw-fw; if (preg2[fw-1]) ans+=2*(cnt2[i-1]-cnt2[fw-1]); else ans+=2*(i-fw-cnt2[i-1]+cnt2[fw-1]); int x=preg2[i-1]^preg2[fw-1]^1; if (x==preg[i-1]) ans+=2*(cnt[lw-1]-cnt[i-1]); else ans+=2*(lw-i-cnt[lw-1]+cnt[i-1]); if (x^preg[i-1]^preg[lw-1]) ans--; ret[i]=min(ret[i],ans); } } } void solve() { scanf("%d",&n); scanf("%s",s+1); rep(i,1,n+1) g[i]=(s[i]=='1'); gao(g,ret,n); reverse(g+1,g+n+1); gao(g,ret2,n); ll ans=0; rep(i,1,n+1) { ret[i]=min(ret[i],ret2[n+1-i]); ans=(ans+(ll)i*ret[i])%mod; } printf("%lld\n",ans); } int main() { for (scanf("%d",&_);_;_--) solve(); }
K题
1 // K 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i<n;i++) 5 #define per(i,a,n) for (int i=n-1;i>=a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first 10 #define se second 11 #define SZ(x) ((int)(x).size()) 12 typedef vector<int> VI; 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 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 18 // head 19 20 double d; 21 int _,h,m,c,sign; 22 char s[20]; 23 int main() { 24 for (scanf("%d",&_);_;_--) { 25 scanf("%d%d%s",&h,&m,s); 26 h=h*60+m; 27 sign=s[3]=='+'?1:-1; 28 sscanf(s+4,"%lf",&d); 29 c=(int)(d*10+0.1); 30 c=sign*c*6-8*60; 31 h+=c; 32 h%=(24*60); 33 if (h<0) h+=24*60; 34 printf("%02d:%02d\n",h/60,h%60); 35 } 36 }
