题解:
https://files.cnblogs.com/files/clrs97/2018CCPC%E5%A5%B3%E7%94%9F%E8%B5%9Bsol.pdf
Code:
1001. CCPC直播
#include<cstdio> #include<cstring> int Case,n,i,x;char s[100]; int main(){ scanf("%d",&Case); while(Case--){ scanf("%s",s); n=strlen(s); for(i=0;i<3-n;i++)putchar(' '); printf("%s|",s); scanf("%s",s); n=strlen(s); printf("%s",s); for(i=0;i<16-n;i++)putchar(' '); scanf("%s",s); printf("|%s|[",s); scanf("%s",s); n=strlen(s); if(s[0]=='R'&&s[1]=='u'){ scanf("%d",&x); for(i=0;i<x;i++)putchar('X'); for(i=0;i<10-x;i++)putchar(' '); }else if(s[0]=='F'){ printf(" AC* "); }else{ printf(" %s",s); for(i=0;i<6-n;i++)putchar(' '); } puts("]"); } }
1002. 口算训练
#include<cstdio> #include<vector> using namespace std; const int N=100010; int Case,n,m,i,x,L,R;vector<int>v[N]; inline void add(int n,int x){ for(int i=2;i*i<=n;i++)while(n%i==0)v[i].push_back(x),n/=i; if(n>1)v[n].push_back(x); } inline bool ask(int x,int k){ int l=0,r=v[x].size()-1,t=-1,mid; while(l<=r)if(v[x][mid=(l+r)>>1]>=L)r=(t=mid)-1;else l=mid+1; if(t<0)return 0; r=v[x].size()-1; while(k--){ if(t>r)return 0; if(v[x][t]>R)return 0; t++; } return 1; } inline bool solve(int n){ for(int i=2;i*i<=n;i++)if(n%i==0){ int t=0; while(n%i==0)n/=i,t++; if(!ask(i,t))return 0; } if(n>1)return ask(n,1); return 1; } int main(){ scanf("%d",&Case); while(Case--){ scanf("%d%d",&n,&m); for(i=2;i<N;i++)v[i].clear(); for(i=1;i<=n;i++)scanf("%d",&x),add(x,i); while(m--)scanf("%d%d%d",&L,&R,&x),puts(solve(x)?"Yes":"No"); } }
1003. 缺失的数据范围
#include<cstdio> typedef long long ll; int Case,a,b;ll K,l,r,mid,ans; inline ll mul(ll a,ll b){ if(!a||!b)return 0; if(a>(K+5)/b)return K+5; a*=b; return a>K+5?K+5:a; } inline ll getpow(ll a,int b){ ll t=1; while(b--)t=mul(t,a); return t; } inline int getlog(ll n){ ll o=n; while(o%2==0)o/=2; int t=0; while(n>1)n/=2,t++; if(o>1)t++; return t; } inline ll cal(ll n,int a,int b){ return mul(getpow(n,a),getpow(getlog(n),b)); } int main(){ scanf("%d",&Case); while(Case--){ scanf("%d%d%lld",&a,&b,&K); l=2,r=K,ans=1; while(l<=r){ mid=(l+r)>>1; if(cal(mid,a,b)<=K)l=(ans=mid)+1;else r=mid-1; } printf("%lld ",ans); } }
1004. 寻宝游戏
#include<cstdio> #include<algorithm> using namespace std; const int N=55,M=25; int Case,n,m,K,i,j,k,x,y,z,a[N][N],f[N][N][M][M],ans,s[N],cnt;//in out inline bool cmp(int x,int y){return x>y;} inline void up(int&a,int b){if(a<b)a=b;} int main(){ scanf("%d",&Case); while(Case--){ scanf("%d%d%d",&n,&m,&K); for(i=1;i<=n;i++)for(j=1;j<=m;j++)scanf("%d",&a[i][j]); for(i=1;i<=n;i++)for(j=1;j<=m;j++)for(x=0;x<=K;x++)for(y=0;y<=K;y++)f[i][j][x][y]=-1; up(f[1][1][0][0],a[1][1]); up(f[1][1][0][1],0); for(i=1;i<=n;i++)for(j=1;j<=m;j++){ if(i<n){ cnt=0; for(k=1;k<=K;k++)s[k]=0; for(k=j+1;k<=m;k++)s[++cnt]=a[i][k]; for(k=1;k<j;k++)s[++cnt]=a[i+1][k]; sort(s+1,s+cnt+1,cmp); for(k=1;k<=K;k++)s[k]+=s[k-1]; } for(x=0;x<=K;x++)for(y=0;y<=K;y++)if(~f[i][j][x][y]){ z=f[i][j][x][y]; //move right if(j<m){ //count in up(f[i][j+1][x][y],z+a[i][j+1]); //swap out if(y<K)up(f[i][j+1][x][y+1],z); } //move down if(i<n)for(k=0;x+k<=K;k++){ //count in up(f[i+1][j][x+k][y],z+s[k]+a[i+1][j]); //swap out if(y<K)up(f[i+1][j][x+k][y+1],z+s[k]); } } } for(ans=x=0;x<=K;x++)up(ans,f[n][m][x][x]); printf("%d ",ans); } }
1005. 奢侈的旅行
#include<cstdio> #include<queue> #include<vector> using namespace std; typedef long long ll; typedef pair<ll,int>P; const int N=100010,M=200010; const ll inf=1LL<<60; int Case,n,m,i,g[N],v[M],a[M],b[M],nxt[M],ed;ll d[N]; priority_queue<P,vector<P>,greater<P> >q; inline void add(int x,int y,int A,int B){v[++ed]=y;a[ed]=A;b[ed]=B;nxt[ed]=g[x];g[x]=ed;} int getlog(ll x){ int t=0; while(x>1)x/=2,t++; return t; } inline void ext(int x,ll y){if(d[x]>y)q.push(P(d[x]=y,x));} inline bool check(ll lv,ll a,ll b){return a/lv>=((1LL<<b)-1);} int main(){ scanf("%d",&Case); while(Case--){ scanf("%d%d",&n,&m); for(ed=0,i=1;i<=n;i++)g[i]=0; while(m--){ int x,y,a,b; scanf("%d%d%d%d",&x,&y,&a,&b); add(x,y,a,b); } for(i=1;i<=n;i++)d[i]=inf; ext(1,1); while(!q.empty()){ P t=q.top();q.pop(); if(d[t.second]<t.first)continue; for(i=g[t.second];i;i=nxt[i])if(check(t.first,a[i],b[i]))ext(v[i],t.first+a[i]); } if(d[n]==inf)puts("-1");else printf("%d ",getlog(d[n])); } }
1006. 对称数
#include<cstdio> #include<cstdlib> #include<algorithm> using namespace std; typedef unsigned long long ll; const int N=200010,M=N*20; int Case,n,m,i,x,y,z,a[N],g[N],v[N<<1],nxt[N<<1],ed; int size[N],son[N],d[N],f[N],top[N],tot,root[N],l[M],r[M]; ll sum[M],ran[N],pre[N]; inline void add(int x,int y){v[++ed]=y;nxt[ed]=g[x];g[x]=ed;} int ins(int x,int a,int b,int c,ll d){ int y=++tot; sum[y]=sum[x]^d; if(a==b)return y; int mid=(a+b)>>1; if(c<=mid){ l[y]=ins(l[x],a,mid,c,d); r[y]=r[x]; }else{ l[y]=l[x]; r[y]=ins(r[x],mid+1,b,c,d); } return y; } void dfs(int x,int y){ f[x]=y;d[x]=d[y]+1; size[x]=1;son[x]=0; root[x]=ins(root[y],1,N,a[x],ran[a[x]]); for(int i=g[x];i;i=nxt[i])if(v[i]!=y){ dfs(v[i],x); size[x]+=size[v[i]]; if(size[v[i]]>size[son[x]])son[x]=v[i]; } } void dfs2(int x,int y){ top[x]=y; if(son[x])dfs2(son[x],y); for(int i=g[x];i;i=nxt[i])if(v[i]!=f[x]&&v[i]!=son[x])dfs2(v[i],v[i]); } inline int lca(int x,int y){ while(top[x]!=top[y]){ if(d[top[x]]<d[top[y]])swap(x,y); x=f[top[x]]; } return d[x]<d[y]?x:y; } inline int ask(int A,int B,int C,int D){ int a=1,b=N; while(a<b){ int mid=(a+b)>>1; if((sum[l[A]]^sum[l[B]]^sum[l[C]]^sum[l[D]])!=(pre[mid]^pre[a-1])){ b=mid; A=l[A]; B=l[B]; C=l[C]; D=l[D]; }else{ a=mid+1; A=r[A]; B=r[B]; C=r[C]; D=r[D]; } } return a; } int main(){ for(i=1;i<N;i++)ran[i]=ran[i-1]*233+17; for(i=1;i<N;i++)pre[i]=pre[i-1]^ran[i]; scanf("%d",&Case); while(Case--){ scanf("%d%d",&n,&m); for(tot=ed=0,i=1;i<=n;i++)g[i]=0; for(i=1;i<=n;i++)scanf("%d",&a[i]); for(i=1;i<n;i++)scanf("%d%d",&x,&y),add(x,y),add(y,x); dfs(1,0); dfs2(1,1); while(m--){ scanf("%d%d",&x,&y); z=lca(x,y); printf("%d ",ask(root[x],root[y],root[z],root[f[z]])); } } }
1007. 赛题分析
#include<cstdio> #include<algorithm> using namespace std; int T,x,y,n,m,i,A,B; int main(){ scanf("%d",&T); for(x=1;x<=T;x++){ scanf("%d%d",&n,&m); A=B=~0U>>1; for(i=1;i<=n;i++){ scanf("%d",&y); A=min(A,y); } for(i=1;i<=m;i++){ scanf("%d",&y); B=min(B,y); } printf("Problem %d: ",x+1000); printf("Shortest judge solution: %d bytes. ",A); if(m)printf("Shortest team solution: %d bytes. ",B); else puts("Shortest team solution: N/A bytes."); } }
1008. quailty算法
#include<cstdio> const int N=300010,inf=~0U>>1; int Case,n,i,a[N],b[N];long long ans; void solve(int o,int l,int r){ if(o<0||l>=r)return; int i,j,L=l-1,R=r+1; for(i=l;i<=r;i++)if(a[i]>>o&1)b[++L]=a[i];else b[--R]=a[i]; for(i=l;i<=r;i++)a[i]=b[i]; solve(o-1,l,L); solve(o-1,R,r); int cl=L-l+1,cr=r-R+1; if(!cl||!cr)return; if(cl>=3&&cr>=3)return; int k=1,A=inf,B=inf; if(cl<3&&cr<3)k++; for(i=l;i<=L;i++)for(j=R;j<=r;j++){ int t=a[i]^a[j]; if(t<A)B=A,A=t; else if(t<B)B=t; } while(k--){ if(A==inf)return; ans+=A; A=B; B=inf; } } int main(){ scanf("%d",&Case); while(Case--){ scanf("%d",&n); for(i=1;i<=n;i++)scanf("%d",&a[i]); ans=0; solve(30,1,n); printf("%lld ",ans); } }
1009. SA-IS后缀数组
#include<cstdio> const int N=1000010; int Case,n,i;char a[N],f[N]; int main(){ scanf("%d",&Case); while(Case--){ scanf("%d%s",&n,a+1); f[n]='>'; for(i=n-1;i;i--){ if(a[i]<a[i+1])f[i]='<'; else if(a[i]>a[i+1])f[i]='>'; else f[i]=f[i+1]; } for(i=1;i<n;i++)putchar(f[i]); puts(""); } }
1010. 回文树
#include<cstdio> #include<algorithm> #include<vector> using namespace std; typedef pair<int,int>P; const int N=100010; int Case,n,i,j,x,y,a[N],ans;vector<P>g[N]; void dfs(int x,int fx,int y,int fy){ if(a[x]!=a[y])return; if(fx&&x==y)return; if(x<=y)ans++; for(int i=0,j=0,k=0;i<g[x].size();i++)if(g[x][i].second!=fx){ while(j<g[y].size()&&g[y][j].first<g[x][i].first)j++; while(k<g[y].size()&&g[y][k].first<=g[x][i].first)k++; for(int o=j;o<k;o++)if(g[y][o].second!=fy)dfs(g[x][i].second,x,g[y][o].second,y); } } int main(){ scanf("%d",&Case); while(Case--){ scanf("%d",&n); for(i=1;i<=n;i++)scanf("%d",&a[i]),g[i].clear(); for(i=1;i<n;i++){ scanf("%d%d",&x,&y); g[x].push_back(P(a[y],y)); g[y].push_back(P(a[x],x)); } for(i=1;i<=n;i++)sort(g[i].begin(),g[i].end()); ans=0; for(i=1;i<=n;i++)dfs(i,0,i,0); for(i=1;i<=n;i++)for(j=0;j<g[i].size();j++)dfs(i,g[i][j].second,g[i][j].second,i); printf("%d ",ans); } }
1011. 代码派对
#include<cstdio> typedef long long ll; const int N=100010,M=1010; int Case,n,m,i,j,e[N][4],s[M][M];ll C3[N],ans; ll solve(int dx,int dy){ for(i=1;i<=m;i++)for(j=1;j<=m;j++)s[i][j]=0; for(i=1;i<=n;i++){ int xl=e[i][0]+dx,yl=e[i][1]+dy,xr=e[i][2],yr=e[i][3]; if(xl>xr||yl>yr)continue; s[xl][yl]++; s[xr+1][yl]--; s[xl][yr+1]--; s[xr+1][yr+1]++; } ll ret=0; for(i=1;i<=m;i++)for(j=1;j<=m;j++){ s[i][j]+=s[i-1][j]+s[i][j-1]-s[i-1][j-1]; ret+=C3[s[i][j]]; } return ret; } int main(){ for(i=1;i<N;i++)C3[i]=1LL*i*(i-1)*(i-2)/6; scanf("%d",&Case); m=1000; while(Case--){ scanf("%d",&n); for(i=1;i<=n;i++)for(j=0;j<4;j++)scanf("%d",&e[i][j]); ans=solve(0,0)-solve(1,0)-solve(0,1)+solve(1,1); printf("%lld ",ans); } }