标签:alarm amp tin sum 约数个数 lun city n+1 parent
A. Survival Route
留坑。
B. Dispersed parentheses
$f[i][j][k]$表示长度为$i$,未匹配的左括号数为$j$,最多的未匹配左括号数为$k$的方案数。时间复杂度$O(n^3)$。
#include<cstdio> #include<algorithm> using namespace std; typedef long long ll; const int P=1000000009; const int N=310; int n,m,i,j,k,f[N][N][N]; inline void up(int&a,int b){a+=b;if(a>=P)a-=P;} int main(){ scanf("%d%d",&n,&m); f[0][0][0]=1; for(i=1;i<=n;i++)for(j=0;j<=i;j++)for(k=j;k<=m;k++){ if(f[i-1][j][k]){ up(f[i][j][k],f[i-1][j][k]); up(f[i][j+1][max(j+1,k)],f[i-1][j][k]); if(j)up(f[i][j-1][k],f[i-1][j][k]); } } printf("%d",f[n][0][m]); }
C. Chocolate triangles
留坑。
D. LWDB
把树的点分治过程记录下来,每个分治结构按覆盖距离维护一个栈,查询时二分即可。时间复杂度$O(n\log^2n)$。
#include<cstdio> #include<algorithm> #include<vector> using namespace std; typedef pair<int,int>P; typedef pair<int,P>PI; const int N=100010,M=3000000; int n,m,i,x,y,z,op; int g[N],nxt[N<<1],v[N<<1],w[N<<1],ok[N<<1],ed; int son[N],f[N],all,now,cnt,value[N]; int G[N],NXT[M],V[M],W[M],ED; vector<PI>q[N]; int top[N]; int Time; //top dis is low but time is new inline void add(int x,int y,int z){ v[++ed]=y; w[ed]=z; nxt[ed]=g[x]; ok[ed]=1; g[x]=ed; } inline void ADD(int x,int y,int w){ V[++ED]=y; W[ED]=w; NXT[ED]=G[x]; G[x]=ED; } void findroot(int x,int y){ son[x]=1,f[x]=0; for(int i=g[x];i;i=nxt[i])if(ok[i]&&v[i]!=y){ findroot(v[i],x); son[x]+=son[v[i]]; if(son[v[i]]>f[x])f[x]=son[v[i]]; } if(all-son[x]>f[x])f[x]=all-son[x]; if(f[x]<f[now])now=x; } void dfs(int x,int y,int dis){ ADD(x,now,dis); for(int i=g[x];i;i=nxt[i])if(ok[i]&&v[i]!=y)dfs(v[i],x,dis+w[i]); } void solve(int x){ int i; dfs(x,0,0); for(i=g[x];i;i=nxt[i])if(ok[i]){ ok[i^1]=0; f[0]=all=son[v[i]]; findroot(v[i],now=0); solve(now); } } inline void paint(int x,int y,int z){ Time++; for(int i=G[x];i;i=NXT[i]){ int w=y-W[i]; if(w<0)continue; int u=V[i]; while(top[u]){ if(w>=q[u][top[u]-1].first)top[u]--; else break; } PI t(w,P(Time,z)); if(top[u]==q[u].size())q[u].push_back(t);else q[u][top[u]]=t; top[u]++; } } inline int query(int x){ P ret(0,0); for(int i=G[x];i;i=NXT[i]){ int w=W[i]; int u=V[i]; if(!top[u])continue; if(q[u][0].first<w)continue; int l=0,r=top[u]-1,mid,fin; while(l<=r){ mid=(l+r)>>1; if(q[u][mid].first>=w)l=(fin=mid)+1;else r=mid-1; } ret=max(ret,q[u][fin].second); } return ret.second; } int main(){ scanf("%d",&n); for(ed=i=1;i<n;i++){ scanf("%d%d%d",&x,&y,&z); add(x,y,z); add(y,x,z); } f[0]=all=n; findroot(1,now=0); solve(now); scanf("%d",&m); while(m--){ scanf("%d%d",&op,&x); if(op==1)scanf("%d%d",&y,&z),paint(x,y,z); else printf("%d\n",query(x)); } }
E. Pea-City
求出凸包之后旋转卡壳。
#include<cstdio> #include<cmath> #include<algorithm> #include<vector> using namespace std; typedef double DB; const int N=88888; const DB eps=1e-8,pi=acos(-1); DB ans; int n; struct PT{ DB x,y; PT(DB x=0,DB y=0):x(x),y(y){} void input(){scanf("%lf%lf",&x,&y);} bool operator<(const PT&p)const{ if(fabs(x-p.x))return x<p.x; return y<p.y; } void output(){printf("%.10f %.10f\n",x,y);} }p[N],q[N]; vector<PT>ret; DB vect(PT p,PT p1,PT p2){ return (p1.x-p.x)*(p2.y-p.y)-(p1.y-p.y)*(p2.x-p.x); } int convex_hull(PT*p,int n,PT*q){ int i,k,m; sort(p,p+n); m=0; for(i=0;i<n;q[m++]=p[i++])while(m>1&&vect(q[m-2],q[m-1],p[i])<eps)m--; k=m; for(i=n-2;i>=0;q[m++]=p[i--])while(m>k&&vect(q[m-2],q[m-1],p[i])<eps)m--; return --m; } PT get(PT p,DB x){ return PT(p.x*cos(x)-p.y*sin(x),p.x*sin(x)+p.y*cos(x)); } bool is_ext(int id,PT pp){ if(vect(p[id],PT(p[id].x+pp.x,p[id].y+pp.y),p[id+1])<-eps)return 0; if(vect(p[id],PT(p[id].x+pp.x,p[id].y+pp.y),p[(id-1+n)%n])<-eps)return 0; return 1; } PT inter(PT p1,PT p2,PT p3,PT p4){ p2.x+=p1.x; p2.y+=p1.y; p4.x+=p3.x; p4.y+=p3.y; DB s=vect(p1,p2,p3),s1=vect(p1,p2,p4); DB t=s/(s-s1); return PT(p3.x+(p4.x-p3.x)*t,p3.y+(p4.y-p3.y)*t); } void solve(){ int f[4]; f[1]=f[2]=f[3]=0; for(int i=0;i<n;i++){ f[0]=i; PT v[4]; v[0]=PT(p[i+1].x-p[i].x,p[i+1].y-p[i].y); for(int j=1;j<4;j++)for(v[j]=get(v[0],pi/2*j);!is_ext(f[j],v[j]);f[j]=(f[j]+1)%n); vector<PT>tmp; for(int j=0;j<4;j++)tmp.push_back(inter(p[f[j]],v[j],p[f[(j+1)%4]],v[(j+1)%4])); DB tmps=0; for(int j=0;j<4;j++)tmps+=vect(tmp[0],tmp[j],tmp[(j+1)%4]); tmps=fabs(tmps); if(ans>tmps)ans=tmps,ret=tmp; } } int main(){ scanf("%d",&n); for(int i=0;i<n;i++)p[i].input(); n=convex_hull(p,n,q); for(int i=0;i<n;i++)p[i]=q[i]; p[n]=p[0]; ans=1e100; solve(); for(int i=0;i<4;i++)ret[i].output(); return 0; }
F. Beautiful sums
等价于求约数个数为$n$的最小奇数,$f[i][j]$表示$i$个质因子,约数个数为$j$的最小奇数,然后DP即可。时间复杂度$O(n\log n)$。
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int Maxn=100020; typedef vector<LL>vi; const double Inf=1e80; const int mod=1e9+9; double dp[17][Maxn]; int pre[17][Maxn],pe[17][Maxn]; int n; vector<int>ys; vector<int>pri; bool isp[100]; int powmod(int x,int y){ int ret=1; while(y){ if(y&1)ret=1LL*ret*x%mod; y>>=1; x=1LL*x*x%mod; } return ret; } int main(){ for(int i=2;i<100;i++){ if(!isp[i])pri.push_back(i); for(int j=i+i;j<100;j+=i)isp[j]=1; } while(scanf("%d",&n)!=EOF){ if(n==1){puts("1");continue;} for(int i=1;i<=n;i++){ if(n%i==0)ys.push_back(i); } for(int i=0;i<=16;i++){ for(int j=1;j<=n;j++)dp[i][j]=Inf; } dp[0][1]=0; for(int i=1;i<=16;i++){ for(int j=0;j<ys.size();j++){ int x=ys[j]; dp[i][x]=Inf; for(int k=0;k<=j;k++){ int y=ys[k]; if(x%y)continue; if(dp[i-1][x/y]+(y-1)*log(pri[i]+.0)<dp[i][x]){ dp[i][x]=dp[i-1][x/y]+(y-1)*log(pri[i]+.0); pre[i][x]=x/y; pe[i][x]=y-1; } } } } vector<int>res; int cur=n; for(int i=16;i>=1;i--){ //printf("cur=%d\n",cur); if(pe[i][cur]>=1)res.push_back(pe[i][cur]); cur=pre[i][cur]; } sort(res.begin(),res.end(),greater<int>()); int ans=1; for(int i=0;i<res.size();i++){ //printf("res=%d\n",res[i]); ans=1LL*ans*powmod(pri[i+1],res[i])%mod; } printf("%d\n",ans); } }
G. Nano alarm-clocks
按题意模拟即可。
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int Maxn=100020; const LL t0=1000000000000LL,t1=1000000; int n; LL x[Maxn]; int main(){ while(scanf("%d",&n)!=EOF){ LL ans=5e18; LL totsum=0; for(int i=1;i<=n;i++){ LL a,b,c;scanf("%lld%lld%lld",&a,&b,&c); x[i]=a*t0+b*t1+c; totsum+=x[i]; } sort(x+1,x+n+1); LL cur=0; LL All=12*t0; for(int i=1;i<=n;i++){ cur+=x[i]; LL bef=i*x[i]-cur; LL aft=(x[i]+All)*(n-i)-(totsum-cur); ans=min(ans,bef+aft); } printf("%lld %lld %lld\n",ans/t0,(ans/t1)%t1,ans%t1); } }
H. Lunch
题意有毒,留坑。
I. Accounting Numeral System
二分然后暴力算组合数,注意要用实数。
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef vector<LL>vi; const int Maxn=2020; LL dp[Maxn][Maxn],sum[Maxn][Maxn]; LL n,m; int ans[Maxn]; void calsum(int idx){ for(int i=0;i<=n;i++){ sum[idx][i]=dp[idx][i]; if(i)sum[idx][i]+=sum[idx][i-1]; } } LL cal(LL x){ if(x<m)return 0; LL tmp=1;//C(1000,1000) if(m+m<=x){ for(LL i=0;i<m;i++){ //if(tmp/(i+1)>ned/(x-i)+1)return 0; //if((long double)ned*(i+1.)+10<(long double)tmp*(x-i))return 0; tmp=tmp*(x-i)/(i+1); } } else{ for(LL i=x;i>m;i--){ //printf("tmp%lld\n",tmp); //if((long double)ned*(x-i+1)+10<(long double)tmp*(i))return 0; tmp=tmp*i/(x-i+1); } } return tmp; } bool check(LL x,LL ned){ if(x<m)return 1; LL tmp=1;//C(1000,1000) if(m+m<=x){ for(LL i=0;i<m;i++){ //if(tmp/(i+1)>ned/(x-i)+1)return 0; if((long double)ned*(i+1.)+10<(long double)tmp*(x-i))return 0; tmp=tmp*(x-i)/(i+1); } } else{ for(LL i=x;i>m;i--){ //printf("tmp%lld\n",tmp); if((long double)ned*(x-i+1)+10<(long double)tmp*(i))return 0; tmp=tmp*i/(x-i+1); } } //printf("tmp=%lld\n",tmp); if(tmp>ned)return 0; return 1; } LL solve(LL ned){ LL l=0,r=1e9; while(l+1<r){ LL mid=(l+r)>>1; if(check(mid,ned))l=mid; else r=mid; // printf("l=%lld r=%lld\n",l,r); } return l; } int main(){ //n=10000000000000LL; //m=2; //printf("%lld",cal(4472136LL)); //solve(7937589951629LL); //printf("%d\n",check(n/2,7937589951629LL)); //m=10; ///printf("%d\n",check(10,1)); while(scanf("%lld%lld",&n,&m)!=EOF){ //solve(n); //printf("%d\n",check(7,n)); int tot=m; LL pre=1e9; for(int i=1;i<=tot;i++){ LL tmp=solve(n); //printf("tmp=%lld\n",tmp); tmp=min(tmp,pre-1); pre=tmp; n-=cal(tmp); printf("%lld%c",tmp,i==tot?‘\n‘:‘ ‘); //printf("tmp=%lld n=%lld\n",tmp,n); m--; } //printf("n=%lld\n",n); } }
J. Ceizenpok’s formula
将模数分解质因数之后递归计算,然后用CRT合并即可。
#include<cstdio> typedef long long ll; ll n,m,x,y,P,B,s[1111111]; ll exgcd(ll a,ll b){ if(!b)return x=1,y=0,a; ll d=exgcd(b,a%b),t=x; return x=y,y=t-a/b*y,d; } ll rev(ll a,ll P){exgcd(a,P);while(x<0)x+=P;return x%P;} ll pow(ll a,ll b,ll P){ ll t=1; for(;b;b>>=1LL,a=a*a%P)if(b&1LL)t=t*a%P; return t; } struct Num{ ll a,b; Num(){a=1,b=0;} Num(ll _a,ll _b){a=_a,b=_b;} Num operator*(Num x){return Num(a*x.a%P,b+x.b);} Num operator/(Num x){return Num(a*rev(x.a,P)%P,b-x.b);} }; Num cal(ll n){return n?Num(s[n%P]*pow(s[P],n/P,P)%P,n/B)*cal(n/B):Num(1,0);} void pre(){ ll i; for(i=s[0]=1;i<P;i++)if(i%B)s[i]=s[i-1]*i%P;else s[i]=s[i-1]; s[P]=s[P-1]; } ll solve(int _B,int _P){ B=_B,P=_P; pre(); Num t=cal(n)/cal(m)/cal(n-m); return 1LL*t.a*pow(B,t.b,P)%P; } ll a[11111],b[11111];int cnt; void divide(int P){ for(int i=2;;i++)if(P%i==0){ int x=1; while(P%i==0)P/=i,x*=i; a[cnt]=x; b[cnt]=solve(i,x); cnt++; if(P==1)return; } } ll CRT(int n){ ll ans=0,P=1; for(int i=0;i<n;i++)P*=a[i]; for(int i=0;i<n;i++)ans=(ans+(P/a[i])*rev(P/a[i],a[i])%P*b[i]%P)%P; return (ans%P+P)%P; } int main(){ int P; scanf("%lld%lld%d",&n,&m,&P); divide(P); printf("%lld",CRT(cnt)); }
K. Dividing an orange
留坑。
L. The Pool for Lucky Ones
按题意模拟即可。
#include<cstdio> #include<algorithm> using namespace std; typedef long long ll; int n,i,a[200000],v[1000000];ll ans=1LL<<60; void change(int x,int a,int b,int c,int p){ v[x]+=p; if(a==b)return; int mid=(a+b)>>1; if(c<=mid)change(x<<1,a,mid,c,p); else change(x<<1|1,mid+1,b,c,p); } inline void upd(){ int a=0,b=100010,mid,x=1; while(a<b){ mid=(a+b)>>1; if(v[x<<1|1])a=mid+1,x=x<<1|1;else b=mid,x<<=1; } ans=min(ans,1LL*a*v[x]); } int main(){ scanf("%d",&n); for(i=1;i<=n;i++)scanf("%d",&a[i]); for(i=1;i<=n;i++)change(1,0,100010,a[i],1); upd(); for(i=1;i<n;i++){ if(a[i]){ change(1,0,100010,a[i],-1); change(1,0,100010,a[i]-1,1); change(1,0,100010,a[i+1],-1); change(1,0,100010,a[i+1]+1,1); upd(); change(1,0,100010,a[i],1); change(1,0,100010,a[i]-1,-1); change(1,0,100010,a[i+1],1); change(1,0,100010,a[i+1]+1,-1); } } for(i=2;i<=n;i++){ if(a[i]){ change(1,0,100010,a[i],-1); change(1,0,100010,a[i]-1,1); change(1,0,100010,a[i-1],-1); change(1,0,100010,a[i-1]+1,1); upd(); change(1,0,100010,a[i],1); change(1,0,100010,a[i]-1,-1); change(1,0,100010,a[i-1],1); change(1,0,100010,a[i-1]+1,-1); } } printf("%lld",ans); }
XV Open Cup named after E.V. Pankratiev. GP of Tatarstan
标签:alarm amp tin sum 约数个数 lun city n+1 parent
原文地址:http://www.cnblogs.com/clrs97/p/6147743.html