Задача А.

#include<bits/stdc++.h>
#define int long long
 using namespace std;

 int a,b,n,k,h,x,y,L,R,m,len,i,l,r,c[100],LL[100],ans,pos,ANSWER,st[100],val,kol;

 inline int f(long long x)
 {
     len = 0;
      while(x>0)
      {
          x/=10;
          ++len;
      }
      return len;
 }

 inline void solve()
 {

     cin>>a>>b>>n>>k;

     x=y=0;

     h = a;
      while(h>0)
      {
        ++x;
        h/=10;
      }

    h = a + b * n;
     while(h>0)
     {
        ++y;
        h/=10;
     }

    L = 0;
    R = n;

     for(i=x;i<=y;++i)
     {
       l = L;
       r = R;
        while(l+1<r)
        {
            m = (l+r)>>1;
            if(f(a + m * b)>i) r=m; else l=m;
        }
        for(m=min(r+10,R);m>=max(L,r-10);--m)
            if(f(a + m * b)==i)
        {
            c[i] = (m-L+1);
            LL[i] = L;
            L=m+1;
            break;
        }
        if(L>R)break;
     }

  ans = 0;

     for(i=x;i<=y;++i)
     {
         ans+=c[i] * i;
         if(ans>=k)
         {
             ans-=c[i] * i;
             k-=ans;
              pos = k / i;
              if(k % i == 0)
              {
                val = a + b * (LL[i] + pos - 1);
                ANSWER=val % 10;
              } else
              {
                val = a + b * (LL[i] + pos);
                k%=i;
                kol = 0;
                while(val>0)
                {
                    st[++kol] = val % 10;
                    val/=10;
                }
                reverse(st+1,st+kol+1);
                ANSWER=st[k];
              }
              cout<<ANSWER<<'\n';
              return;
         }
     }

     cout<<-1<<'\n';
 }

   main()
  {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

      solve();

    return 0;
  }

 

Задача В.

#include<bits/stdc++.h>
 using namespace std;

 int m[] = {31,28,31,30,31,30,31,31,30,31,30,31};
 pair<int,int> add[100],val[100][100],ans;
 int t,i,j,pos,x,y,n;
 string s;

  int main()
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    cin>>t;
    m[1]+=t;
    cin>>n;
    add[0] = {1000,0};
    add[1] = {5000,0};
    add[2] = add[3] = {3000,0};
    add[4] = {0,3};

    for(i=1;i<=12;++i)
        for(j=1;j<=m[i-1];++j)
    {
        val[j][i] = add[pos];
        ++pos;
        if(pos==5) pos = 0;
    }

    for(i=1;i<=n;++i)
    {
        cin>>s;
        x =  (s[0] - '0') * 10 + (s[1] - '0');
        y = (s[3] - '0') * 10 + (s[4] - '0');
        ans.first+=val[x][y].first;
        ans.second+=val[x][y].second;
    }

    cout<<ans.first<<' '<<ans.second<<'\n';

  }

 

Задача С.

#include<bits/stdc++.h>
 using namespace std;

 int x,y,n,k,tt,i;
 set<pair<int,int> > st;

  int main()
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

     cin>>n;
      for(i=1;i<=n;++i)
      {
        cin>>x>>y;
        if(x<0)tt=1,x=-x; else tt = 0;
         k = __gcd(x,y);
         x/=k;
         y/=k;
         if(tt)x=-x;
         st.insert({x,y});
      }

      cout<<(int)st.size()<<'\n';

  }

 

Задача D.

#include<bits/stdc++.h>
#define N 1111
#define left jdjdjs
#define right sjssjsjj
using namespace std;

int up[N][N],left[N][N],down[N][N],right[N][N],l[N],r[N],n,m,ptr,mink,ans,kol;
char c[N][N];
int s[N];

pair<int,int> times[N];

inline int prev(int x){
        return x&(x-1);
}

inline int next(int x){
        return (x<<1)-(x&(x-1));
}

inline void modify(int pos,int value){
        for (;pos<N;pos=next(pos)) s[pos]+=value;
}

inline int get_sum(int l,int r){
        if (l>r) return 0;
        int k=0,p;
        for (p=r;p>0;p=prev(p)) k+=s[p];
        for (p=l-1;p>0;p=prev(p)) k-=s[p];
        return k;
}

inline void input(){
        scanf("%d%d%d\n",&n,&m,&mink);
        mink=mink%4?mink/4+2:mink/4+1;
        for (int i=1;i<=n;i++){
                for (int j=1;j<=m;j++) scanf("%c",&c[i][j]);
                scanf("\n");
        }
}

inline void init(){
        for (int i=1;i<=n;i++)
                for (int j=1;j<=m;j++){
                        up[i][j]=c[i][j]=='#'?0:up[i-1][j]+1;
                        left[i][j]=c[i][j]=='#'?0:left[i][j-1]+1;
                }
        for (int i=n;i>=1;i--)
                for (int j=m;j>=1;j--){
                        down[i][j]=c[i][j]=='#'?0:down[i+1][j]+1;
                        right[i][j]=c[i][j]=='#'?0:right[i][j+1]+1;
                }
}

inline void upd_ans(){
        memset(s,0,sizeof(s));
        kol=0;
        for (int i=1;i<=ptr;i++){
                if (l[i]>=mink){
                        modify(i,1);
                        times[++kol]=make_pair(l[i]+i-1,i);
                }
        }
        sort(times+1,times+kol+1);
        int f=1;
        for (int i=1;i<=ptr;i++){
                ans+=get_sum(i-r[i]+1,i-mink+1);
                while (times[f].first==i && f<=kol){
                        modify(times[f].second,-1);
                        ++f;
                }
        }
}

int main(){
        input();
        init();
        for (int I=1;I<=n;I++){
                ptr=0;
                for (int i=I,j=1;i<=n && j<=m;i++,j++){
                        l[++ptr]=min(right[i][j],down[i][j]);
                        r[ptr]=min(left[i][j],up[i][j]);
                }
                upd_ans();
        }
        for (int I=2;I<=m;I++){
                ptr=0;
                for (int i=1,j=I;i<=n && j<=m;i++,j++){
                        l[++ptr]=min(right[i][j],down[i][j]);
                        r[ptr]=min(left[i][j],up[i][j]);
                }
                upd_ans();
        }
        printf("%d",ans);
}

 

Задача Е.

#include <iostream>
#include <cstdio>
#include <vector>
#include <map>
#include <algorithm>
#include <ctime>

using namespace std;

struct query {
    int parent;
    int color;
    int index;

    query(int p, int c, int i):
        parent(p), color(c), index(i) {}

    bool operator < (const query& other) const {
        if (parent != other.parent)
            return parent < other.parent;
        return color < other.color;
    }
};

void solve() {
    int N, M, K;
    scanf("%d%d%d", &N, &M, &K);

    vector<int> A(N + 1);
    for (int i = 1; i <= N; ++i)
        scanf("%d", &A[i]);

    vector<int> P(N + 1);
    P[2] = 1;
    for (int i = 3; i <= N; ++i)
        scanf("%d", &P[i]);

    vector<int> X(M), Y(M);
    for (int mi = 0; mi < M; ++mi) {
        scanf("%d%d", &X[mi], &Y[mi]);
    }

    vector<query> colors;
    vector<int> p_index(N + 1);
    vector<int> v_index(N + 1);
    vector<int> value(N + N + M + M);
    vector<int> p_new_index(M);
    vector<int> v_new_index(M);

    colors.reserve(N + N + M + M);
    for (int i = 2; i <= N; ++i) {
        colors.push_back(query(P[i], A[i], i));
    }
    for (int i = 1; i <= N; ++i) {
        colors.push_back(query(i, A[i], i + N));
    }

    for (int mi = 0; mi < M; ++mi) {
        colors.push_back(query(P[X[mi]], Y[mi], 2 * N + mi + 1));
        colors.push_back(query(X[mi], Y[mi], 2 * N + M + mi + 1));
    }

    sort(colors.begin(), colors.end());

    int ind = 0;
    for (int i = 0; i < colors.size();) {
        int j;
        for (j = i; j < colors.size() && colors[i].parent == colors[j].parent && colors[i].color == colors[j].color; ++j) {
            if (colors[j].index <= N) {
                p_index[colors[j].index] = ind;
                ++value[ind];
                continue;
            }
            if (colors[j].index <= 2 * N) {
                v_index[colors[j].index - N] = ind;
                continue;
            }
            if (colors[j].index <= 2 * N + M) {
                p_new_index[colors[j].index - 1 - N - N] = ind;
                continue;
            }
            v_new_index[colors[j].index - 1 - N - N - M] = ind;
        }
        i = j;
        ++ind;
    }

    int ans = 0;
    for (int i = 1; i <= N; ++i)
        ans += value[v_index[i]];

    for (int mi = 0; mi < M; ++mi) {
        if (A[P[X[mi]]] == A[X[mi]]) --ans;
        ans -= value[v_index[X[mi]]];
        --value[p_index[X[mi]]];

        A[X[mi]] = Y[mi];

        if (A[P[X[mi]]] == A[X[mi]]) ++ans;
        p_index[X[mi]] = p_new_index[mi];
        v_index[X[mi]] = v_new_index[mi];

        ans += value[v_index[X[mi]]];
        ++value[p_index[X[mi]]];

        printf("%d\n", ans);
    }
}

int main() {
    solve();

    cerr << clock() << endl;
    return 0;
}