Educational Codeforces Round 8

开始填坑_(:з」∠)_

628A - Tennis Tournament    20171124

小学数学题，\((x,y)=((n-1)\cdot(2b+1),np)\)

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,b,p,x,y;
int main()
{
    scanf("%d%d%d",&n,&b,&p);
    x=(n-1)*(2*b+1),y=n*p;
    printf("%d %d\n",x,y);
    return 0;
}

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628B - New Skateboard    20171124

由于能根据末尾的两个数字判断出是否为4的倍数，直接每相邻两位判断一下就好了

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
string s;
LL ans;
bool check(int k){return k%4==0;}
int main()
{
    cin>>s;
    ans=check(s[0]-'0');
    for(LL i=1;i<s.size();i++)
      ans+=check(s[i]-'0'),ans+=i*check(s[i-1]*10+s[i]);
    printf("%I64d\n",ans);
    return 0;
}

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628C - New Skateboard    20171124

让每一位修改产生的距离尽可能大即可

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
string s;
int n,k,f[100001];
int main()
{
    scanf("%d%d",&n,&k);
    cin>>s;
    for(int i=1;i<=n;i++)
      f[i]=max(s[i-1]-'a','z'-s[i-1])+f[i-1];
    if(f[n]<k)return printf("-1\n"),0;
    for(int i=0;i<n;i++)
      {
      if(k>=f[i+1]-f[i])
        s[i]=(s[i]-'a'>'z'-s[i])?'a':'z';
      else s[i]=(s[i]-k>='a')?s[i]-k:s[i]+k;
      k-=f[i+1]-f[i];if(k<=0)break;
      }
    cout<<s<<endl;return 0;
}

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628D - Magic Numbers    20190301

比较恶心的DP，首先将区间\((l,r)\)的询问转化为两次对区间\((1,n)\)的询问

令\(f[i][j][k]\)表示当前进行到第\(i\)位，除以\(m\)的余数为\(j\)时，与\(n\)的大小关系为\(k\)的方案数

循环时\(i\)从\(1\)到\(n\)，通过枚举第\(i\)位的数字来推出\(j\)进行转移

接下去就是各种细节处理了

#include<bits/stdc++.h>
using namespace std;
#define N 2005
#define LL long long
#define MOD 1000000007
LL m,d,f[N][N][3],ans;
char s[N];
LL rua()
{
    LL n=strlen(s+1);
    memset(f,0,sizeof(f));
    for(LL j=1;j<=9;j++)if(j!=d)
      {
      if(j<s[1]-'0')f[1][j%m][0]++;
      if(j==s[1]-'0')f[1][j%m][1]++;
      if(j>s[1]-'0')f[1][j%m][2]++;
      }
    for(LL i=1;i<n;i++)
      for(LL j=0;j<m;j++)
        for(LL k=0;k<3;k++)if(f[i][j][k])
          {
          //cout<<i<<" "<<j<<" "<<k<<" "<<f[i][j][k]<<endl;
          if(i&1)
            {
            if(k==0)(f[i+1][(j*10ll+d)%m][k]+=f[i][j][k])%=MOD;
            if(k==1)
              {
              if(d<s[i+1]-'0')(f[i+1][(j*10ll+d)%m][0]+=f[i][j][k])%=MOD;
              else if(d==s[i+1]-'0')(f[i+1][(j*10ll+d)%m][1]+=f[i][j][k])%=MOD;
              else if(d>s[i+1]-'0')(f[i+1][(j*10ll+d)%m][2]+=f[i][j][k])%=MOD;
              }
            if(k==2)(f[i+1][(j*10ll+d)%m][k]+=f[i][j][k])%=MOD;
            continue;
            }
          for(LL l=0;l<=9;l++)if(l!=d)
            {
            if(k==0)(f[i+1][(j*10ll+l)%m][k]+=f[i][j][k])%=MOD;
            if(k==1)
              {
              if(l<s[i+1]-'0')(f[i+1][(j*10ll+l)%m][0]+=f[i][j][k])%=MOD;
              else if(l==s[i+1]-'0')(f[i+1][(j*10ll+l)%m][1]+=f[i][j][k])%=MOD;
              else if(l>s[i+1]-'0')(f[i+1][(j*10ll+l)%m][2]+=f[i][j][k])%=MOD;
              }
            if(k==2)(f[i+1][(j*10ll+l)%m][k]+=f[i][j][k])%=MOD;
            }
          }
    LL res=0;
    //for(LL j=0;j<m;j++)for(LL k=0;k<2;k++)if(f[n][j][k])cout<<n<<" "<<j<<" "<<k<<" "<<f[n][j][k]<<endl;
    for(LL i=1;i<n;i++)
      for(LL k=0;k<3;k++)
        (res+=f[i][0][k])%=MOD;
    (res+=f[n][0][0]+f[n][0][1])%=MOD;
    return res;
}
bool check()
{
    LL n=strlen(s+1);
    for(LL i=1;i<=n;i++)
      {
      if((i&1) && s[i]==d+'0')
        return false;
      if(i%2==0 && s[i]!=d+'0')
        return false;
      }
    LL r=0;
    for(LL i=1;i<=n;i++)
      r=(r*10+s[i]-'0')%m;
    return r==0;
}
int main()
{
    scanf("%I64d%I64d",&m,&d);
    scanf("%s",s+1);
    ans=MOD-rua()+check();
    //cout<<rua()<<endl;
    scanf("%s",s+1);
    ans+=rua();
    //cout<<rua()<<endl;
    return printf("%I64d\n",ans%MOD),0;
}

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628E - Zbazi in Zeydabad    20190305

这题只能说。。。bitset大法好，莽就完事了

#include<bits/stdc++.h>
using namespace std;
#define N 3005
#define LL long long
LL n,m,ans;
bitset<N>a[N],b[N],c[N];
int get()
{
    char ch=getchar();
    while(ch!='.' && ch!='z')ch=getchar();
    return ch=='z';
}
int main()
{
    scanf("%I64d%I64d",&n,&m);
    for(LL i=0;i<n;i++)
      for(LL j=0;j<m;j++)
        a[i][m-j-1]=get(),b[i].set(),c[i].set();
    for(LL i=0;i<min(m,n);i++)
      {
      for(LL j=0;j<n;j++)b[j]&=a[j]>>i;
      for(LL j=0;j<n-i;j++)c[j]&=a[j+i]>>i,ans+=(b[j]&b[j+i]&c[j]).count();
      }
    return printf("%I64d\n",ans),0;
}

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628F - Bear and Fair Set    20190306

标解是网络流，但是有种瞎搞的方法不知道为什么能过_(:з」∠)_

考虑\(\{0,1,2,3,4\}\)的所有子集\(A\)，对每个区间，计算出满足\(i\%5\epsilon A\)的\(i\)的个数，并与该区间可取数字的个数取\(min\)，加起来后若个数小于\(\frac{|A|\cdot n}{5}\)，则一定无解

若考虑完所有子集后依旧没有出现不合法的情况，则一定有解（本弱不会证，求大佬证明）

#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
int nn,n,b,q;
vector<pair<int,int> >d;
int main()
{
    d.push_back(mp(0,0));
    scanf("%d%d%d",&n,&b,&q),nn=n;
    d.push_back(mp(b,n));
    for(int i=1;i<=q;i++)scanf("%d%d",&b,&n),d.push_back(mp(b,n));
    sort(d.begin(),d.end());
    for(int i=1;i<=q;i++)
       if(d[i].second>d[i+1].second || d[i+1].second-d[i].second>d[i+1].first-d[i].first)
         return printf("unfair\n"),0;
    for(int k=0;k<32;k++)
      {
      int res=0;
      for(int i=1;i<=q+1;i++)
        {
        int cnt=0;
        for(int j=d[i-1].first+1;j<=d[i].first;j++)
          cnt+=bool((1<<(j%5))&k);
        res+=min(cnt,d[i].second-d[i-1].second);
        }
      if(res<nn/5*__builtin_popcount(k))return printf("unfair\n"),0;
      }
    printf("fair\n");
    return 0;
}

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