A. An Olympian Math Problem

  1.  #include <cstdio>
  2.  #include <cstdlib>
  3.  #include <cmath>
  4.  #include <cstring>
  5.  #include <time.h>
  6.  #include <string>
  7.  #include <set>
  8.  #include <map>
  9.  #include <list>
  10.  #include <stack>
  11.  #include <queue>
  12.  #include <vector>
  13.  #include <bitset>
  14.  #include <ext/rope>
  15.  #include <algorithm>
  16.  #include <iostream>
  17.  using namespace std;
  18.  #define ll long long
  19.  #define minv 1e-6
  20.  #define inf 1e9
  21.  #define pi 3.1415926536
  22.  #define nl 2.7182818284
  23.  const ll mod=1e9+;//
  24.  const int maxn=1e5+;
  25.  
  26.  int main()
  27.  {
  28.      int t;
  29.      ll n;
  30.      scanf("%d",&t);
  31.      while (t--)
  32.      {
  33.          scanf("%lld",&n);
  34.          printf("%lld\n",n-);
  35.      }
  36.      return ;
  37.  }

B. The writing on the wall

与 https://leetcode.com/problems/maximal-rectangle/description/ 这道题很像

  1.  #include <bits/stdc++.h>
  2.  using namespace std;
  3.  #define ll long long
  4.  const ll mod=1e9+;
  5.  const int maxn=1e5+;
  6.  const int maxm=1e2+;
  7.  
  8.  int a[maxn][maxm],c[maxn][maxm];
  9.  int qx[maxm],qy[maxm];
  10.  
  11.  int main()
  12.  {
  13.      ll sum=;
  14.      int t,T,n,m,g,x,y,i,j,k;
  15.      bool vis;
  16.      scanf("%d",&t);
  17.      for (T=;T<=t;T++)
  18.      {
  19.          memset(a,,sizeof(a));
  20.          scanf("%d%d%d",&n,&m,&g);
  21.          while (g--)
  22.          {
  23.              scanf("%d%d",&x,&y);
  24.              a[x][y]=;
  25.          }
  26.          for (j=;j<=m;j++)
  27.              for (i=;i<=n;i++)
  28.                  c[i][j]=(a[i][j]==)?:c[i-][j]+;
  29.          sum=;
  30.          for (i=;i<=n;i++)
  31.          {
  32.              ///以a[i][j]作为右下方
  33.              g=;
  34.              for (j=;j<=m;j++)
  35.              {
  36.                  if (g== || c[i][j]>qx[g])
  37.                      vis=;
  38.                  else
  39.                      vis=;
  40.                  sum+=c[i][j];
  41.                  qy[g+]=j;
  42.  
  43.                  while (qx[g]>c[i][j])
  44.                      sum+=1ll*(qy[g+]-qy[g])*c[i][j],g--;
  45.                  k=g;
  46.                  while (k)
  47.                      sum+=1ll*(qy[k+]-qy[k])*qx[k],k--;
  48.                  if (g== || qx[g]!=c[i][j])
  49.                      g++;
  50.                  if (vis)
  51.                      qy[g]=j;
  52.                  qx[g]=c[i][j];
  53.              }
  54.          }
  55.          printf("Case #%d: %lld\n",T,sum);
  56.      }
  57.      return ;
  58.  }
  59.  /*
  60.  100
  61.  3 3 3
  62.  1 1
  63.  1 2
  64.  2 1
  65.  
  66.  3 3 4
  67.  1 1
  68.  1 2
  69.  2 1
  70.  2 2
  71.  
  72.  2 3 0
  73.  
  74.  100000 100 0
  75.  */

C. GDY

  1.  #include <cstdio>
  2.  #include <cstdlib>
  3.  #include <cmath>
  4.  #include <cstring>
  5.  #include <time.h>
  6.  #include <string>
  7.  #include <set>
  8.  #include <map>
  9.  #include <list>
  10.  #include <stack>
  11.  #include <queue>
  12.  #include <vector>
  13.  #include <bitset>
  14.  #include <ext/rope>
  15.  #include <algorithm>
  16.  #include <iostream>
  17.  using namespace std;
  18.  #define ll long long
  19.  #define minv 1e-6
  20.  #define inf 1e9
  21.  #define pi 3.1415926536
  22.  #define nl 2.7182818284
  23.  const ll mod=1e9+;//
  24.  const int maxn=2e2+;
  25.  const int maxm=2e4;
  26.  
  27.  int s[maxm],a[maxn][maxm],g[maxn];
  28.  int value[]={,,,,,,,,,,,,,};
  29.  
  30.  int cmp(int x,int y)
  31.  {
  32.      return value[x]>value[y];
  33.  }
  34.  
  35.  int main()
  36.  {
  37.      int t,T,n,m,num,i,j,sum;
  38.      int x,y,z;
  39.      scanf("%d",&t);
  40.      for (T=;T<=t;T++)
  41.      {
  42.          memset(g,,sizeof(g));
  43.  
  44.          scanf("%d%d",&n,&m);
  45.          for (i=;i<m;i++)
  46.              scanf("%d",&s[i]);
  47.          num=;
  48.          for (i=;i<n;i++)
  49.          {
  50.              for (j=;j<;j++)
  51.              {
  52.                  if (num==m)
  53.                      break;
  54.                  a[i][j]=s[num++];
  55.              }
  56.              sort(a[i],a[i]+j,cmp);
  57.              g[i]=j;
  58.          }
  59.  
  60.          z=a[][g[]-];//previous number
  61.          g[]--;
  62.          y=;//has y persons
  63.          x=;//pos
  64.          while ()
  65.          {
  66.              for (i=g[x]-;i>=;i--)
  67.                  if (value[z]+==value[a[x][i]] || (a[x][i]== && z!=))
  68.                      break;
  69.  
  70.              if (i!=-)
  71.              {
  72.                  z=a[x][i];
  73.                  a[x][i]=;
  74.                  sort(a[x],a[x]+g[x],cmp);
  75.                  g[x]--;
  76.                  if (g[x]==)
  77.                      break;
  78.                  y=;
  79.              }
  80.              else if (y!=n-)
  81.                  y++;
  82.              else
  83.              {
  84.                  x=(x+)%n;
  85.                  for (i=x;;i=(i+)%n)
  86.                  {
  87.                      if (num==m)
  88.                          break;
  89.                      else
  90.                      {
  91.                          a[i][g[i]++]=s[num++];
  92.                          sort(a[i],a[i]+g[i],cmp);
  93.                      }
  94.                      if (i==(x-+n)%n)
  95.                          break;
  96.                  }
  97.  
  98.                  z=a[x][g[x]-];
  99.                  g[x]--;
  100.                  if (g[x]==)
  101.                      break;
  102.                  y=;
  103.              }
  104.  
  105.              x=(x+)%n;
  106.          }
  107.          printf("Case #%d:\n",T);
  108.          for (i=;i<n;i++)
  109.              if (i==x)
  110.                  printf("Winner\n");
  111.              else
  112.              {
  113.                  sum=;
  114.                  for (j=;j<g[i];j++)
  115.                      sum+=a[i][j];
  116.                  printf("%d\n",sum);
  117.              }
  118.      }
  119.      return ;
  120.  }
  121.  /*
  122.  10
  123.  2 6
  124.  3 5 7 9 11 4
  125.  
  126.  3 20
  127.  2 2 2 2 2 2 2 2 2 2
  128.  2 2 2 2 2 2 2 2 2 2
  129.  
  130.  3 19
  131.  2 2 2 2 2 2 2 2 2 2
  132.  2 2 2 2 2 2 2 2 2
  133.  
  134.  3 20
  135.  3 2 2 2 2 2 2 2 2 2 2
  136.  2 2 2 2 2 2 2 2 2
  137.  
  138.  2 10
  139.  3 4 5 6 7 2 3 4 5 6
  140.  
  141.  3 15
  142.  3 4 5 6 7 12 12 12 12 12 13 13 13 13 13
  143.  
  144.  3 11
  145.  1 2 3 4 5 6 7 8 9 10 11
  146.  
  147.  3 13
  148.  1 2 3 4 5 6 7 8 9 10 11 12 13
  149.  
  150.  1
  151.  3 17
  152.  1 2 3 4 5 6 7 8 9 10 11 12 13 12 11 10 9
  153.  
  154.  */

E. AC Challenge

  1. #include <cstdio>
  2. #include <cstdlib>
  3. #include <cmath>
  4. #include <cstring>
  5. #include <time.h>
  6. #include <string>
  7. #include <set>
  8. #include <map>
  9. #include <list>
  10. #include <stack>
  11. #include <queue>
  12. #include <vector>
  13. #include <bitset>
  14. #include <ext/rope>
  15. #include <algorithm>
  16. #include <iostream>
  17. using namespace std;
  18. #define ll long long
  19. #define minv 1e-6
  20. #define inf -1e18
  21. #define pi 3.1415926536
  22. #define nl 2.7182818284
  23. const ll mod=1e9+;//
  24. const int maxn=1e5+;
  25.  
  26. ll f[<<];
  27.  
  28. struct node
  29. {
  30.     ll value;
  31.     int pos,t;
  32. };
  33.  
  34. struct cmp1
  35. {
  36.     bool operator() (node a,node b)
  37.     {
  38.         return a.value>b.value;
  39.     }
  40. };
  41.  
  42. //priority_queue<node>st;
  43. priority_queue<node,vector<node>,cmp1>st;
  44. int er[],a[],b[],v[];
  45.  
  46. int main()
  47. {
  48.     int n,m,s,i,j,value,pos,t;
  49.     ll r=,rr;
  50.     for (i=;i<;i++)
  51.         er[i]=(<<i);
  52.  
  53.     scanf("%d",&n);
  54.     for (i=;i<(<<n);i++)
  55.             f[i]=inf;
  56.  
  57.     for (i=;i<n;i++)
  58.     {
  59.         scanf("%d%d%d",&a[i],&b[i],&m);
  60.         v[i]=;
  61.         while (m--)
  62.         {
  63.             scanf("%d",&s);
  64.             v[i]+=(er[s-]);
  65.         }
  66.         if (v[i]==)
  67.         {
  68.             f[er[i]]=a[i]+b[i];
  69.             st.push({a[i]+b[i],er[i],});
  70.             r=max(r,(ll)a[i]+b[i]);
  71.         }
  72.     }
  73.  
  74.     while (!st.empty())
  75.     {
  76.         value=st.top().value;
  77.         pos=st.top().pos;
  78.         t=st.top().t;
  79.         st.pop();
  80.         for (i=;i<n;i++)
  81.             if ((pos & er[i])== && (pos & v[i])==v[i] && value+1ll*t*a[i]+b[i]>f[pos|er[i]])
  82.             {
  83.                 rr=value+1ll*t*a[i]+b[i];
  84.                 f[pos|er[i]]=rr;
  85.                 r=max(r,rr);
  86.                 st.push({rr,pos|er[i],t+});
  87.             }
  88.     }
  89.     printf("%lld",r);
  90.     return ;
  91. }
  92. /*
  93. 3
  94. 1 2 0
  95. 1 5 0
  96. 8 1 0
  97. */

G. Lpl and Energy-saving Lamps

  1.  #include <cstdio>
  2.  #include <cstdlib>
  3.  #include <cmath>
  4.  #include <cstring>
  5.  #include <time.h>
  6.  #include <string>
  7.  #include <set>
  8.  #include <map>
  9.  #include <list>
  10.  #include <stack>
  11.  #include <queue>
  12.  #include <vector>
  13.  #include <bitset>
  14.  #include <ext/rope>
  15.  #include <algorithm>
  16.  #include <iostream>
  17.  using namespace std;
  18.  #define ll long long
  19.  #define minv 1e-6
  20.  #define inf 1e18
  21.  #define pi 3.1415926536
  22.  #define nl 2.7182818284
  23.  const ll mod=1e9+;//
  24.  const int maxn=1e5+;
  25.  
  26.  struct node
  27.  {
  28.      int x,y;
  29.  }f[maxn],r[maxn];
  30.  
  31.  int tag[maxn<<],a[maxn],x,y;
  32.  
  33.  int cmp(node a,node b)
  34.  {
  35.      return a.x<b.x;
  36.  }
  37.  
  38.  void build(int index,int l,int r)
  39.  {
  40.      if (l==r)
  41.          scanf("%d",&tag[index]);
  42.      else
  43.      {
  44.          int m=(l+r)>>;
  45.          build(index<<,l,m);
  46.          build(index<<|,m+,r);
  47.          tag[index]=min(tag[index<<],tag[index<<|]);
  48.      }
  49.  }
  50.  
  51.  int query(int index,int l,int r,int v)
  52.  {
  53.      if (l==r)
  54.      {
  55.          if (tag[index]>v)
  56.              return ;
  57.  
  58.          x++;
  59.          y-=tag[index];
  60.          tag[index]=inf;
  61.          return l;
  62.      }
  63.      else
  64.      {
  65.          int m=(l+r)>>,z;
  66.          if (tag[index<<]<=v)
  67.              z=query(index<<,l,m,v);
  68.  
  69.          else
  70.              z=query(index<<|,m+,r,v);
  71.          tag[index]=min(tag[index<<],tag[index<<|]);
  72.          return z;
  73.      }
  74.  }
  75.  
  76.  int main()
  77.  {
  78.      int n,m,q,Q,d,index,i,j;
  79.      scanf("%d%d",&n,&m);
  80.      build(,,n);
  81.  
  82.      scanf("%d",&q);
  83.      for (Q=;Q<=q;Q++)
  84.      {
  85.          scanf("%d",&d);
  86.          f[Q].x=d;
  87.          f[Q].y=Q;
  88.      }
  89.      sort(f+,f+q+,cmp);
  90.  
  91.      index=;
  92.      x=;
  93.      y=;
  94.      for (i=;i<=f[q].x;i++)
  95.      {
  96.          y+=m;
  97.          j=;
  98.          while (j)
  99.              j=query(,,n,y);
  100.  
  101.          while (f[index].x==i)
  102.          {
  103.              r[f[index].y]={x,y};
  104.              index++;
  105.          }
  106.      }
  107.  
  108.      for (i=;i<=q;i++)
  109.          printf("%d %d\n",r[i].x,r[i].y);
  110.      return ;
  111.  }

J. Sum

  1.  #include <cstdio>
  2.  #include <cstdlib>
  3.  #include <cmath>
  4.  #include <cstring>
  5.  #include <time.h>
  6.  #include <string>
  7.  #include <set>
  8.  #include <map>
  9.  #include <list>
  10.  #include <stack>
  11.  #include <queue>
  12.  #include <vector>
  13.  #include <bitset>
  14.  #include <ext/rope>
  15.  #include <algorithm>
  16.  #include <iostream>
  17.  using namespace std;
  18.  #define ll long long
  19.  #define minv 1e-6
  20.  #define inf 1e9
  21.  #define pi 3.1415926536
  22.  #define nl 2.7182818284
  23.  const ll mod=1e9+;//
  24.  const int maxn=2e7+;
  25.  
  26.  int zhi[maxn],f[maxn];
  27.  bool vis[maxn],v[maxn];
  28.  
  29.  int main()
  30.  {
  31.      int t,i,j,k,x=,y=,value=2e7;
  32.      ll n,sum;
  33.      memset(vis,,sizeof(vis));
  34.      memset(v,,sizeof(v));
  35.      for (i=;i<=value;i++)
  36.      {
  37.          if (!vis[i])
  38.          {
  39.              x++;
  40.              zhi[x]=i;
  41.          }
  42.          for (j=;j<=x;j++)
  43.          {
  44.              k=i*zhi[j];
  45.              if (k>value)
  46.                  break;
  47.              vis[k]=;
  48.              v[k]=v[i];
  49.              if (i%zhi[j]==)
  50.              {
  51.                  v[k]=;
  52.                  break;
  53.              }
  54.          }
  55.      }
  56.      for (i=;i<=value;i++)
  57.          if (!v[i])
  58.          {
  59.              y++;
  60.              f[y]=i;
  61.          }
  62.  
  63.  //    for (i=1;i<=100;i++)
  64.  //        printf("%d ",f[i]);
  65.  
  66.      scanf("%d",&t);
  67.      f[]=;
  68.      while (t--)
  69.      {
  70.          scanf("%lld",&n);
  71.          sum=;
  72.          j=y;
  73.          for (i=;i<=y;i++)
  74.          {
  75.              while (1ll*f[i]*f[j]>n)
  76.                  j--;
  77.              sum+=j;
  78.          }
  79.          printf("%lld\n",sum);
  80.      }
  81.      return ;
  82.  }

L. Magical Girl Haze

  1.  /*
  2.  图问题:
  3.  spfa容易被卡时间复杂度
  4.  
  5.  而dijkstra是贪心，不会被卡
  6.  */
  7.  #include <cstdio>
  8.  #include <cstdlib>
  9.  #include <cmath>
  10.  #include <cstring>
  11.  #include <time.h>
  12.  #include <string>
  13.  #include <set>
  14.  #include <map>
  15.  #include <list>
  16.  #include <stack>
  17.  #include <queue>
  18.  #include <vector>
  19.  #include <bitset>
  20.  #include <ext/rope>
  21.  #include <algorithm>
  22.  #include <iostream>
  23.  using namespace std;
  24.  #define ll long long
  25.  #define minv 1e-6
  26.  #define inf 1e18
  27.  #define pi 3.1415926536
  28.  #define nl 2.7182818284
  29.  const ll mod=1e9+;//
  30.  const int maxn=1e5+;
  31.  
  32.  struct rec
  33.  {
  34.      int d,len;
  35.      rec *to;
  36.  }*e[maxn];
  37.  
  38.  struct node
  39.  {
  40.      ll dist;
  41.      int k,d;
  42.  };
  43.  
  44.  struct cmp1
  45.  {
  46.      bool operator() (node a,node b)
  47.      {
  48.          if (a.dist==b.dist)
  49.              return a.k>b.k;
  50.          else
  51.              return a.dist>b.dist;
  52.      }
  53.  };
  54.  
  55.  priority_queue<node,vector<node>,cmp1>st;
  56.  
  57.  ll f[maxn][];
  58.  
  59.  int main()
  60.  {
  61.      rec *p;
  62.      int t,n,m,k,i,j,d,kk,dd,x,y,z;
  63.      ll dist,r;
  64.      scanf("%d",&t);
  65.      while (t--)
  66.      {
  67.          scanf("%d%d%d",&n,&m,&k);
  68.          for (i=;i<=n;i++)
  69.              e[i]=NULL;
  70.          while (m--)
  71.          {
  72.              scanf("%d%d%d",&x,&y,&z);
  73.              p=(rec*) malloc (sizeof(rec));
  74.              p->d=y;
  75.              p->len=z;
  76.              p->to=e[x];
  77.              e[x]=p;
  78.          }
  79.          for (i=;i<=n;i++)
  80.              for (j=;j<=k;j++)
  81.                  f[i][j]=inf;
  82.          for (j=;j<=k;j++)
  83.          {
  84.              f[][j]=;
  85.              st.push({,j,});
  86.          }
  87.          //f[i][j]作为一个状态，100000*10
  88.          while (!st.empty())
  89.          {
  90.              dist=st.top().dist;
  91.              d=st.top().d;
  92.              kk=st.top().k;
  93.              st.pop();
  94.              p=e[d];
  95.              while (p)
  96.              {
  97.                  dd=p->d;
  98.                  if (f[dd][kk]>dist+p->len)
  99.                  {
  100.                      f[dd][kk]=dist+p->len;
  101.                      st.push({f[dd][kk],kk,dd});
  102.                  }
  103.                  if (kk!=k && f[dd][kk+]>dist)
  104.                  {
  105.                      f[dd][kk+]=dist;
  106.                      st.push({f[dd][kk+],kk+,dd});
  107.                  }
  108.                  p=p->to;
  109.              }
  110.          }
  111.          r=inf;
  112.          for (j=;j<=k;j++)
  113.              r=min(r,f[n][j]);
  114.          printf("%lld\n",r);
  115.      }
  116.      return ;
  117.  }

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