Possible with doubles, but quite complicated. Easy exact solution is possible with rational numbers (fractions), e.g., in the second case it is 390/43.

cin.sync_with_stdio(false);

Ignore the notes. G++ is faster than Clang++, while Visual C++ 2017 gives TL.

3 line python code

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

#define re return
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sqrt(x) sqrt(abs(x))
#define mp make_pair
#define pi (3.14159265358979323846264338327950288419716939937510)
#define fo(i, n) for(int i = 0; i < n; ++i)
#define ro(i, n) for(int i = n - 1; i >= 0; --i)
#define unique(v) v.resize(unique(all(v)) - v.begin())

template <class T> T abs (T x) { re x > 0 ? x : -x; }
template <class T> T sqr (T x) { re x * x; }
template <class T> T gcd (T a, T b) { re a ? gcd (b % a, a) : b; }
template <class T> int sgn (T x) { re x > 0 ? 1 : (x < 0 ? -1 : 0); }


typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef double D;
typedef long double ld;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef unsigned long long ull;

vi l, r;

int main() {
    //freopen("input.txt", "r", stdin);
    string str;
    while (getline(cin, str)) {
        l.clear(), r.clear();
        if (isalpha(str[0])) l.pb(0);
        fo(i, (int)str.size() - 1) {
            if (!isalpha(str[i]) && isalpha(str[i + 1])) l.pb(i + 1);
            if (isalpha(str[i]) && !isalpha(str[i + 1])) r.pb(i + 1);
        }
        if (isalpha(str.back())) r.pb(str.size());
        fo(i, l.size()) reverse(str.begin() + l[i], str.begin() + r[i]);
        cout << str << '\n';
        str.clear();
    }
}

What is maximum number of the rectangles? According to my idea this is equal 6. I found the surfaces of these rectangles but Wrong answer #5.

Edited by author 20.11.2016 19:45

Here is my code:
# include <bits/stdc++.h>

using namespace std;

string s[300];

string conv(int x) {
    string str;
    while(x) {
        str.push_back(x % 10 + '0');
        x /= 20;
    }
    reverse(str.begin(), str.end());
    return str;
}

int main() {
    int n;
    cin >> n;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= i; j++) {
            s[i] += "sin(" + conv(j);
            if(j != i) {
                if(j % 2)
                    s[i].push_back('-');
                else
                    s[i].push_back('+');
            }
        }
        for(int j = 1; j <= i; j++)
            s[i].push_back(')');
    }

    for(int i = 1; i < n; i++) cout << '(';
    for(int i = 1; i <= n; i++) {
        cout << s[i] << '+' << n - i + 1;
        if(i != n)
            cout << ')';
    }
    cout << endl;
    return 0;
}

Edited by author 13.11.2015 09:37

Edited by author 13.11.2015 09:37

del

Edited by author 11.03.2018 11:13

Edited by author 11.03.2018 11:13

If i use hash-table, i get WA. Because due to the small MOD ~10^6, i get probably a coincidence of some hashes.
(I used the degree = 31 and tried modules 1000003,... etc.)

If i use simply linear hash (in common vector) and sorting i get TL.

If i keep hash of each string in separate vectors, and then use a binary search, then again i get TL.

:(

WA6:
#include <iostream>
#pragma comment(linker, "/STACK:46777216")
#include <vector>
using namespace std;

#define ll int64_t

const int MAXN = 50005;

ll t[4*MAXN], add[4*MAXN], sz[MAXN];
ll n, q, x, cnt;
ll y, z, value[MAXN];
char c[15];
vector <ll> g[MAXN];
ll p[MAXN], pos[MAXN];


void init(ll v, ll tl, ll tr)
{
   if(tl == tr) add[v]=(ll)0, t[v]=value[tl];
   else
   {
      ll m=(tl+tr)/2;
      init(2*v, tl, m);
      init(2*v+1, m+1,tr);
      add[v]=(ll)0, t[v]=t[2*v]+t[2*v+1];
   }
}

ll sum(ll v, ll tl, ll tr, ll l, ll r)
{
   if(tl == l && tr == r) return t[v];
   else
   {
        ll m = (tl + tr) /2 ;
        ll a = add[v] * (ll)(r-l+1);
        if(r<=m) return a+sum(2*v, tl, m, l, r);
        if(l>m) return a+sum(2*v + 1, m+1, tr, l, r);
        return a+sum(2*v, tl, m, l, m) + sum(2*v + 1, m+1, tr, m+1, r);
   }
}


void update(ll v, ll tl, ll tr, ll l, ll r, ll val)
{
   if(tl == l && tr == r) add[v]+=val, t[v]+=val*(ll)(tr-tl+1);
   else
   {
        ll m = (tl + tr) /2;
        if(r<=m) update(2*v , tl, m, l, r, val);
        else
        {
            if(l>m) update(2*v  + 1, m+1, tr, l, r, val);
            else update(2*v , tl, m, l, m, val), update(2*v  + 1, m+1, tr, m+1, r, val);
        }
        t[v]=t[2*v]+t[2*v +1];
   }
};




void dfs(ll v, ll par = -1)
{
    p[v] = par, pos[v] = ++cnt, sz[v] = (ll)1;
    for (ll i = 0; i < (ll) g[v].size(); i++)
    {
        ll to = g[v][i];
        if (to == par) continue;
        dfs(to, v);
        sz[v] += sz[to];
    }
}



int main()
{
    scanf("%lld %lld %lld", &n, &q, &value[1]);
    for (ll i = 2; i <= n; i++)
    {
       scanf("%lld %lld\n", &x, &value[i]);
       x++;
       g[x].push_back(i);
       g[i].push_back(x);
    }

    cnt = 0;
    dfs(1);

    init(1, 1, n);

    for (ll i = 1; i <= q; i++)
    {
        scanf("%s %lld %lld %lld\n",&c, &x, &y, &z);
        x++;
        if (c[0] == 'e')
        {
            ll salary=sum(1,1,n,pos[x],pos[x]);
            if(salary<y) update(1,1,n,pos[x],pos[x],z);
        }
        else
        {
           ll departament = sum(1,1,n,pos[x],pos[x]+sz[x]-1);
           if(departament<y*sz[x]) update(1,1,n,pos[x],pos[x]+sz[x]-1,z);
        }
    }

    for(ll i=1;i<=n;i++)
    {
       ll salary=sum(1,1,n,pos[i],pos[i]);
       printf("%lld", salary);
       if(i<n) printf("\n");
    }

    return 0;
}


WA8:
#include <iostream>
#pragma comment(linker, "/STACK:46777216")
#include <vector>
using namespace std;

#define ll unsigned long long

const int MAXN = 50005;

ll sum[MAXN], add[MAXN], sz[MAXN], L[MAXN], R[MAXN];
int n, q, x, cnt, c;
ll y, z, value[MAXN], len;
char str[15];
vector <int> g[MAXN];
int p[MAXN], pos[MAXN], ind[MAXN];


void Create()
{
   len=(ll)1;
   while(len*len<n) len++;
   if(len*len>n) len--;
   c= n / len;
   for(int i=1;i<=c;i++) L[i]=(ll)(1+len*(i-1)), R[i]=(ll)(len*i);
   if(c*len<n) c++, L[c]=R[c-1]+1, R[c]=n;

   for(int i=1;i<=c;i++)
   {
      sum[i]=(ll)0, add[i]=(ll)0;
      for(int j=L[i];j<=R[i];j++) sum[i]+=value[j], ind[j]=i;
   }
}


void Update(int l, int r, ll val)
{
   int k1=ind[l];
   int k2=ind[r];

   for(int i=k1+1;i<=k2-1;i++) sum[i]+=val*(ll)(R[i]-L[i]+1), add[i]+=val;

   if(k1==k2)
   {
      for(int i=l;i<=r;i++) value[i]+=val, sum[k1]+=val;
   }
   else
   {
      if(l==L[k1]) sum[k1]+=val*(ll)(R[k1]-L[k1]+1), add[k1]+=val;
      else
      {
         for(int i=l;i<=R[k1];i++) value[i]+=val, sum[k1]+=val;
      }

      if(r=R[k2]) sum[k2]+=val*(ll)(R[k2]-L[k2]+1), add[k2]+=val;
      else
      {
         for(int i=L[k2];i<=r;i++) value[i]+=val, sum[k2]+=val;
      }
   }
}


ll Sum(int l, int r)
{
   int k1=ind[l];
   int k2=ind[r];
   ll res=(ll)0;

   for(int i=k1+1;i<=k2-1;i++) res+=sum[i];

   if(k1==k2)
   {
      for(int i=l;i<=r;i++) res+=value[i];
      res+=add[k1]*(ll)(r-l+1);
   }
   else
   {
      for(int i=l;i<=R[k1];i++) res+=value[i];
      res+=add[k1]*(ll)(R[k1]-l+1);

      for(int i=L[k2];i<=r;i++) res+=value[i];
      res+=add[k2]*(ll)(r-L[k2]+1);
   }
   return res;
}


void Dfs(int v, int par = -1)
{
    p[v] = par, pos[v] = ++cnt, sz[v] = (ll)1;
    for (int i = 0; i < (int) g[v].size(); i++)
    {
        int to = g[v][i];
        if (to == par) continue;
        Dfs(to, v);
        sz[v] += sz[to];
    }
}


int main()
{
    scanf("%d%d%lld\n", &n, &q, &value[1]);
    for (int i = 2; i <= n; i++)
    {
       scanf("%d%lld\n", &x, &value[i]);
       x++;
       g[x].push_back(i);
       g[i].push_back(x);
    }

    Dfs(1);
    Create();

    scanf("\n");
    for (int i = 1; i <= q; i++)
    {
        scanf("%s%d%lld%lld\n",&str, &x, &y, &z);
        x++;
        if (str[0] == 'e')
        {
            ll salary=Sum(pos[x],pos[x]);
            if(salary<y) Update(pos[x],pos[x],z);
        }
        else
        {
           ll departament = Sum(pos[x],pos[x]+sz[x]-1);
           if(departament<y*sz[x]) Update(pos[x],pos[x]+sz[x]-1,z);
        }
    }

    for(int i=1;i<=n;i++)
    {
       ll salary=Sum(pos[i],pos[i]);
       printf("%lld", salary);
       if(i<n) printf("\n");
    }

    return 0;
}



Edited by author 10.03.2018 23:40

#include <iostream>
#include <string>

using namespace std;

int main()
{
    int n,m=1;
    string k;
    cin>>n;
    cin>>k;
    int j=k.length();
    for(int i=n;i>=(n%j);i-=j){
        m=m*i;

    }


    cout<<m<<k;



    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    pair<int ,int > a[150000];
    for(int i=0;i<n;i++)
    {
        cin >> a[i].first >> a[i].second;
    }
    for(int i=100;i>=0;--i)
    {
        for(int j=0;j<n;j++)
        {
            if(i==a[j].second) cout << a[j].first << " " << a[j].second << endl;
        }
    }

}

WA #2

a = input().split()
s = input().split()
n, k = int(a[0]), int(a[1])
out = ["" for x in range(n//k+1)]
for i in range(k):
    if i != k - 1:
        for j in range(n//k+1):
            if len(s[0]) == 1:
                out[j] += "   " + s[0]
                s.pop(0)
            elif len(s[0]) == 2:
                out[j] += "  " + s[0]
                s.pop(0)
            else:
                out[j] += " " + s[0]
                s.pop(0)
    else:
        for j in range(n%k):
            if len(s[0]) == 1:
                out[j] += "   " + s[0]
                s.pop(0)
            elif len(s[0]) == 2:
                out[j] += "  " + s[0]
                s.pop(0)
            else:
                out[j] += " " + s[0]
                s.pop(0)
for i in out:
    print(i)

Edited by author 09.03.2018 19:59

It seems that the main problem is precision.
If it is so, how could we compute the answer more accurately?
e.g., set what Epsilon threshold for determining a value being 0 ?

output limit exceeded!
what's the case 4 like?
3Q

#include<stdio.h>
int main()
{
    int n , m , sum=0 ;
    double ava;
    scanf("%d",&n);
    int mark[11];
    for(m=0;m<n;m++){
        scanf("%d",&mark[m]);    }
    for(m=0;m<n;m++){
        sum=sum+mark[m];
    }
    ava=(double)sum/n;
    if(ava<=3)printf("None\n");
    else if (ava>=5)printf("Named\n");
    else if(ava>=4.5)printf("High\n");
    else printf("Common\n");
}