标签：申请 noi 任务 printf 1.0 orm clear back tde
t1
XBG
#include<map> #include<cstdio> #include<string> #include<string.h> #include<iostream> #include<algorithm> using namespace std; int d[123456]; char ty[123456][20]; int n,m; //... int main(){ freopen("toy.in","r",stdin); freopen("toy.out","w",stdout); scanf("%d %d",&n,&m); for(int i=0;i<n;i++)scanf("%d%s",d+i,ty[i]); //for(int i=1;i<=n;i++)d[i]=d[i]?1:-1; int now=0; while(m--){ int op,s; scanf("%d %d",&op,&s); if(!d[now]){ if(op)now=(now+s)%n; else now=((now-s)%n+n)%n; }else{ if(op)now=((now-s)%n+n)%n; else now=(now+s)%n; } } cout<<ty[now]; return 0; }
t2
暴力30
子任务Si=1
以1为根统计每个子树里面1的个数
子任务链的写错了。
45分
#include<map> #include<cstdio> #include<string> #include<vector> #include<string.h> #include<iostream> #include<algorithm> using namespace std; struct edge{ int to,next; }e[601234]; int cnt,last[301234],dep[301234],w[301234],ans[301234]; int n,m; int s[301234],t[301234]; void insert(int a,int b){ e[++cnt]=(edge){b,last[a]};last[a]=cnt; e[++cnt]=(edge){a,last[b]};last[b]=cnt; } int fa[301234][20]; void dfs(int v,int f){ //fa[v][0]=f; //for(int i=1;i<=20;i++)fa[v][i]=fa[fa[v][i-1]][i-1]; for(int i=last[v];i;i=e[i].next){ int b=e[i].to;if(b!=f){ dep[b]=dep[v]+1; dfs(b,v); } } } int an[301234]; void dfs2(int v,int t,int f){ fa[v][0]=f; //for(int i=1;i<=20;i++)fa[v][i]=fa[fa[v][i-1]][i-1]; for(int i=last[v];i;i=e[i].next){ int b=e[i].to;if(b!=f){ dfs2(b,t,v); if(b==t){ an[v]=i; } if(an[b]){ an[v]=i; } } } } void dfs3(int s,int t){ int v=s,ti=0; ans[s]+=(ti==w[s]); while(s!=t){ s=e[an[s]].to; ++ti; ans[s]+=(ti==w[s]); } } int siz[301234]; void dfs4(int v,int f){ //fa[v][0]=f; //for(int i=1;i<=20;i++)fa[v][i]=fa[fa[v][i-1]][i-1]; for(int i=last[v];i;i=e[i].next){ int b=e[i].to;if(b!=f){ dep[b]=dep[v]+1; dfs4(b,v); siz[v]+=siz[b]; //fprintf(stderr,"Debug (%d)=%d\n",v,siz[v]); } } } vector<int>vec[301234]; int main(){ freopen("running.in","r",stdin); freopen("running.out","w",stdout); scanf("%d %d",&n,&m); for(int i=1,u,v;i<n;i++){ scanf("%d %d",&u,&v); insert(u,v); } //dfs(1,0); for(int i=1;i<=n;i++)scanf("%d",w+i); for(int i=1;i<=m;i++)scanf("%d %d",s+i,t+i); if(n<=1000){ for(int i=1;i<=m;i++){ memset(an,0,sizeof(an)); dfs2(s[i],t[i],0); dfs3(s[i],t[i]); } for(int i=1;i<=n;i++)printf("%d ",ans[i]); }else if(n%10==4){ for(int i=1;i<=n;i++)if(t[i]>=s[i])vec[i].push_back(s[i]); for(int i=1;i<=n;i++){ if(i-w[i]>=0){ int Ans=vec[i-w[i]].size(),cnt=0; for(int x=0;x<Ans;x++){ if(t[vec[i-w[i]][x]]<i)cnt++; } ans[i]+=(Ans-cnt); } } //for(int i=1;i<=n;i++)printf("%d ",ans[i]); for(int i=1;i<=n;i++)vec[i].clear(); for(int i=1;i<=n;i++)if(t[i]<s[i])vec[i].push_back(s[i]); for(int i=1;i<=n;i++){ if(i+w[i]<=n){ int Ans=vec[i+w[i]].size(),cnt=0; //fprintf(stderr,"at %d\n",Ans); for(int x=0;x<Ans;x++){ if(t[vec[i+w[i]][x]]>i){ cnt++; } } ans[i]+=(Ans-cnt); } } for(int i=1;i<=n;i++)printf("%d ",ans[i]); }else { for(int i=1;i<=n;i++)siz[t[i]]++; dep[1]=0; dfs4(1,0); for(int i=1;i<=n;i++)printf("%d ",(w[i]==dep[i])*siz[i]); } return 0;//... //... //... }
t3
2^n*2^m暴力枚举所有申请和批准情况
暴力64，数组小了不然是72
要特判m<2
#include<map> #include<cstdio> #include<string> #include<vector> #include<string.h> #include<iostream> #include<algorithm> using namespace std; int n,m,v,e; int c[333],d[333]; double k[333];/* int min(int a,int b){ return a<b?a:b; }*/ struct zy{ int i,cost; }F[1234]; bool cmp(zy a,zy b){ return a.cost>b.cost; } double brutef(){ double ans=1e23; int _F=0; for(int i=1;i<=n;i++){ int tmp=0; if(tmp>1)tmp=tmp+g[c[i-1]][c[i]]-g[c[i-1]][d[i]]; if(tmp<n)tmp=tmp+g[c[i+1]][c[i]]-g[c[i+1]][d[i]]; F[++_F]=(zy){i,-tmp}; } sort(F+1,F+_F+1,cmp); int q[1234]; memset(q,0,sizeof(q)); for(int i=0;i<=m;i++){ q[F[i].i]=1;double tmp2=0; for(int t=1;t<n;t++){ int a=c[t],b=c[t+1]; if(q[t])a=d[t]; if(q[t+1])b=d[t+1]; tmp2+=g[a][b]; } //cout<<F[i].i<<","<<tmp2<<endl; ans=min(ans,tmp2); } return ans; } int main() { freopen("classroom.in","r",stdin); freopen("classroom.out","w",stdout); scanf("%d%d%d%d",&n,&m,&v,&e); for(int i=1;i<=n;i++)scanf("%d",c+i); for(int i=1;i<=n;i++)scanf("%d",d+i); bool xz2=1; for(int i=1;i<=n;i++)scanf("%lf",k+i),xz2=xz2&&(k[i]==1.0); for(int i=1;i<=v;i++)for(int j=1;j<=v;j++)g[i][j]=(i!=j)*(1<<20); for(int i=1;i<=e;i++){ int a,b,w; scanf("%d %d %d",&a,&b,&w); g[a][b]=min(g[a][b],w); g[b][a]=min(g[b][a],w); } for(int p=1;p<=v;p++)for(int i=1;i<=v;i++)for(int j=1;j<=v;j++)g[i][j]=min(g[i][j],g[i][p]+g[p][j]); if(m<2){ double ans=0,tmp=0; for(int i=1;i<n;i++){ ans+=(g[c[i]][c[i+1]]); } tmp=ans; if(m) for(int x=1;x<=n;x++){ //cout<<x<<" :\n"; int res=0; for(int i=1;i<n;i++){ int a=c[i],b=c[i+1]; if(i==x)a=d[x]; if(i+1==x)b=d[x]; res+=g[a][b]; } //cout<<res*k[x]+tmp*(1-k[x])<<endl; ans=min(ans,res*k[x]+tmp*(1-k[x])); } printf("%.2lf",(double)ans); }else if(n<=20&&m<=10){ double ans=1e23; for(int i=0;i<(1<<n);i++){ int p[30],q[33],cnt=0; for(int j=1;j<=n;j++)if(i&(1<<(j-1)))q[++cnt]=j; if(cnt<=m){ //cerr<<"safe"; memset(p,0,sizeof(p)); double tmp=0; for(int j=0;j<(1<<cnt);j++){ double prob=1,tmp2=0; for(int t=1;t<=cnt;t++){ if(j&(1<<(t-1))){ prob=prob*k[q[t]]; p[q[t]]=1; }else { prob=prob*(1-k[q[t]]); p[q[t]]=0; } } /*printf("(%d,%d)\n",i,j); for(int t=1;t<=cnt;t++)cout<<q[t]<<"!!";cout<<endl;*/ for(int t=1;t<n;t++){ int a=c[t],b=c[t+1]; if(p[t])a=d[t]; if(p[t+1])b=d[t+1]; tmp2+=g[a][b]; } //cout<<tmp2<<endl;; tmp+=tmp2*prob; } //cout<<"---"<<tmp<<endl; ans=min(ans,tmp); } }//... printf("%.2lf",(double)ans); }else{ printf("%.2lf",(double)brutef()); } }
t1
XBG
#include<cstdio> #include<algorithm> #include<iostream> using namespace std; int t,k,n,m,c[2333][2333],s[2333][2333]; int main(){ freopen("problem.in","r",stdin); freopen("problem.out","w",stdout); scanf("%d%d",&t,&k); for(int i=0;i<=2000;i++)c[i][1]=i%k; for(int i=1;i<=2000;i++) for(int j=2;j<=i;j++) c[i][j]=(c[i-1][j]+c[i-1][j-1])%k; for(int i=1;i<=2000;i++) for(int j=1;j<=2000;j++){ s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+(j<=i)*(!c[i][j]); } while(t--){ scanf("%d %d",&n,&m); /*for(int i=1;i<=n;i++,puts("")) for(int j=1;j<=i;j++)printf("%d ",s[i][j]); */printf("%d\n",s[n][m]); } return 0; }
t2
拿三个队列，一个存x，一个存px，一个存x-px。
然后这三个队列是单调递减的，具体证明较简单。
#include<queue> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; int n,m,q,u,v,t;double p; int a[123456]; int PQ[123456],_PQ,od[7123456]; namespace solution{ int x[7123456][2],y[7123456][2],z[7123456][2]; int hx,hy,hz,tx,ty,tz; void init(){hx=hy=hz=1;tx=n;ty=tz=0;} void debug(int tim){ for(int i=hx;i<=n;i++)printf("%d ",x[i][0]+q*(tim-x[i][1])); printf("!");for(int i=hy;i<=ty;i++)printf("%d ",y[i][0]+q*(tim-y[i][1])); printf("!");for(int i=hz;i<=tz;i++)printf("%d ",z[i][0]+q*(tim-z[i][1])); printf("!"); puts(""); } //wo hui zuo d2t2 ei void getres(){//debug(m); for(int i=1;i<=n+m;i++){ int c=-12345;int *I; if(hx<=n)if(x[hx][0]+q*(m-x[hx][1])>c)c=x[hx][0]+q*(m-x[hx][1]),I=&hx; if(hy<=ty)if(y[hy][0]+q*(m-y[hy][1])>c)c=y[hy][0]+q*(m-y[hy][1]),I=&hy; if(hz<=tz)if(z[hz][0]+q*(m-z[hz][1])>c)c=z[hz][0]+q*(m-z[hz][1]),I=&hz; ++*I; if(!(i%t))printf("%d ",c); } } int main(){ sort(a+1,a+n+1,greater<int>());//nlogn for(int i=1;i<=n;i++)x[i][0]=a[i]; p=u;p/=v; init(); for(int i=1;i<=m;i++){ int c=-12345; int *I; if(hx<=n)if(x[hx][0]+q*(i-1-x[hx][1])>c)c=x[hx][0]+q*(i-1-x[hx][1]),I=&hx; if(hy<=ty)if(y[hy][0]+q*(i-1-y[hy][1])>c)c=y[hy][0]+q*(i-1-y[hy][1]),I=&hy; if(hz<=tz)if(z[hz][0]+q*(i-1-z[hz][1])>c)c=z[hz][0]+q*(i-1-z[hz][1]),I=&hz; int A,B; A=p*c; B=c-A; //printf("cut %d into(%d,%d)\n",c,A,B); y[++ty][0]=A; z[++tz][0]=B; z[tz][1]=y[ty][1]=i; ++*I; od[i]=c; //cerr<<"h..="<<hx<<endl; //debug(i); } for(int i=t;i<=m;i+=t)printf("%d ",od[i]);puts(""); getres(); return 0; } } int main(){ freopen("earthworm.in","r",stdin); freopen("earthworm.out","w",stdout); //cerr<<sizeof(solution::x)*3; scanf("%d%d%d%d%d%d",&n,&m,&q,&u,&v,&t); for(int i=1;i<=n;i++)scanf("%d",a+i); if(n<=1000&&m<=1000){ for(int i=1;i<=n;i++)PQ[i]=a[i];_PQ=n; for(int _t=1;_t<=m;_t++){ int mx=1; for(int i=2;i<=_PQ;i++)if(PQ[i]>PQ[mx])mx=i; int x=PQ[mx]; PQ[mx]=(int)u*1.0/v*x; PQ[++_PQ]=x-PQ[mx]; //printf("cut %d into(%d,%d)\n",x,PQ[mx],PQ[_PQ]); for(int i=1;i<_PQ;i++)if(i!=mx)PQ[i]+=q; od[_t]=x; } for(int i=t;i<=m;i+=t)printf("%d ",od[i]); sort(PQ+1,PQ+n+m+1,greater<int>()); puts(""); for(int i=t;i<=n+m;i+=t)printf("%d ",PQ[i]); }else solution::main(); return 0; }
t3
枚举两个点作抛物线，预处理能过什么点
状压（即或的值为2^n-1)
注意处理单点，分数未知
riio写错了
总结：d1t2 d2t3暴力写错了很伤
d2t2很像我之前写的QingdaoRegional的G
最后100+45+64+100+100+?未完待续
标签：申请 noi 任务 printf 1.0 orm clear back tde
原文地址：http://www.cnblogs.com/chouti/p/6082750.html
