“字节跳动杯”2018中国大学生程序设计竞赛-女生专场

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\n",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\n",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\n",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\n",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:\n",x+1000);
    printf("Shortest judge solution: %d bytes.\n",A);
    if(m)printf("Shortest team solution: %d bytes.\n",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\n",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\n",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\n",ans);
  }
}

  

posted @ 2018-08-05 18:01 Claris 阅读(...) 评论(...) 编辑 收藏