#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
    int N1, C1, N2, limit, C2, N3, n, time = 0, mins, secs, ans;
    cin >> N1 >> C1    >> N2 >> limit >> C2 >> N3 >> n;
    for (int i = 0; i < n; i++) {
        scanf("%d:%d", &mins, &secs);
        ans = mins * 60 + secs;
        time += ans > 6 ? (ans + 59) / 60 : 0;
    }
    cout << "Basic:     " << N1 + time*C1 << '\n'
         << "Combined:  " << N2 + max(time - limit, 0) * C2 << '\n'
         << "Unlimited: " << N3;
}

Edited by author 25.09.2020 23:01

The cases when all times are less than 6 seconds are neglected. So the cost of each plan should be zero as all calling will be free but adding the basic plan to the cost leads to AC.

You need to remember that string "no more than 6 sec". It means that you need ignoring all calls with 0 min and 0-6 sec including 6 sec. It's important for tests.

//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("avx")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#define re return
#define pb push_back
#define eb emplace_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;
typedef tree <pair<int, char>, null_type, less<pair<int, char>>, rb_tree_tag, tree_order_statistics_node_update> _tree;

int calc () {
    string str;
    cin >> str;
    re (str[0] - '0') * 600 + (str[1] - '0') * 60 + stoi(str.substr(3, 2));
}

int main() {
    int n1, n2, n3, c1, c2, t;
    cin >> n1 >> c1 >> n2 >> t >> c2 >> n3;
    int k, time;
    int ans = 0;
    cin >> k;
    while (k--) {
        time = calc();
        if (time <= 6) continue;
        ans += (time + 59) / 60;
    }
    cout << "Basic:     " << n1 + ans * c1 << endl;
    cout << "Combined:  " << n2 + max(0, ans - t) * c2 << endl;
    cout << "Unlimited: " << n3 << endl;
}

plz can someone give the 4th test

I got WA#6, what is wrong?

can someone tell me the test case 4 for this question. kindly help

#include <iostream>
#include <iomanip>
int main()
{
    unsigned short int p1, c1;
    unsigned short int p2, t, c2;
    unsigned short int p3;
    unsigned short int k;
    unsigned short int mm, ss;
    char s = ':';

    std::cin >> p1 >> c1;
    std::cin >> p2 >> t >> c2;
    std::cin >> p3;
    std::cin >> k;
    for (int i=1; i<=k; i++)
    {
        std::cin >> mm >> s >> ss;
        if (mm>0 || ss>7)
        {
            if (ss!=0)
                mm++;
        p1 = p1 + mm*c1;
        if (mm>t)
        {
            mm-=t;
            t=0;
        }
        else
        {
            t-=mm;
            mm=0;
        }
        p2 = p2 + mm*c2;
        }
    }
    std::cout << setiosflags(std::ios::left) << std::setw(11) << "Basic: " << p1 << std::endl;
    std::cout << setiosflags(std::ios::left) << std::setw(11) << "Combined: " << p2 << std::endl;
    std::cout << setiosflags(std::ios::left) << std::setw(11) << "Unlimited: " << p3 << std::endl;

    return 0;
}


What's wrong?

If your output format differs from the required output format (even by one space).....you will get WA. So be careful.....

If you have WA 10, check your solution on this test:

0 0
0 0 1000
0
2
59:59
00:06

Right answer is:

Basic:     0
Combined:  60000
Unlimited: 0

Edited by author 29.04.2012 16:14

Please,if any one can write all the test case's....

0.001s and 128Kb :)

...
for (i=0; i<num; i++)
    {
       scanf("%d:%d", &min, &sec);
       if (min==0 && sec<7) continue;
       if (sec != 0)
          min += 1;
       basic_price += min*basic_pm;
       min_all += min;
    }
...

p.s.: HF & GL, it's very easy

Can anyone gave me test 8 ?




Edited by author 09.08.2011 16:37

#include <iostream>
#include <stdio.h>

using namespace std;


int main(){
int b[2],c[3],u,n,min,sec,min_all;
cin>>b[0]>>b[1]>>c[0]>>c[1]>>c[2]>>u>>n;
for (int i=0; i < n; i++) {
    scanf("%d:%d", &min, &sec);
    if (min==0 && sec<7) continue;
    if (sec != 0)
    min += 1;
    b[0] += min*b[1];
    min_all += min;
}
printf("Basic: %d\n",b[0]);
printf("Combined: %d\n",c[0]);
printf("Unlimited: %d\n",u);
system("pause");
return 0;
}

WA1 (

#include <stdio.h>
int main()
{
    int n1,c1,n2,t,c2,n3,k,i;
    scanf("%d %d",&n1,&c1);
    scanf("%d %d %d",&n2,&t,&c2);
    scanf("%d",&n3);
    scanf("%d",&k);
    int (*A)[2] = new int[k][2];
    for( i=0; i<k; ++i )
    {
         scanf("%d:%d",&A[i][0],&A[i][1]);
    }
    int basic = n1,combined = n2,unlimited = n3,bas=0,com=0;

    for( i=0; i<k; ++i )
    {
         if(A[i][0]!=0)
         {
                        bas+=A[i][0]; com+=A[i][0];
         }
         if( (A[i][0]==0 && A[i][1]>6) || (A[i][0]!=0 && A[i][1]!=0 ) ) { ++bas; ++com; }
    }
    if ( com>t )
         {
               com-=t; com*=c2;
         }
    bas *= c1;
    basic += bas; combined += com;

    printf("Basic:     %d\n",basic);
    printf("Combined:  %d\n",combined);
    printf("Unlimited: %d\n",unlimited);
    delete [] A;
    return 0;
}

When sum<=T (which means that the limit is not  exceeded ) you should write just a monthly fee N2.

In C++ :
if (sum<=T)
   cout<<"Combined:  "<<N2<<endl;

Good Luck Friends!!!

can somebody please tell me how format of output should be??
thanks..

:) thank you

Edited by author 05.11.2007 17:26

help me please.
Why I've WA 1 ?
I use spaces in output format and my answers are true(((((

var i,k,n1,c1,p,n2,t,c2,n3:word;
    sum,sek:longword;
    code:integer;
    s:string;
begin
  //assign(input,'input.txt');
 // assign(output,'ouput.txt');
  readln(n1,c1);
  readln(n2,t,c2);
  readln(n3);
  readln(k);
  for i:=1 to k do
   begin
    readln(s);
    val(s,p,code);
    val(copy(s,4,2),sek,code);
    if (p=0) and (sek<=6) then sum:=sum
    else
     begin
      sum:=sum+p;
      if sek>0 then sum:=sum+1;
     end;
   end;
  writeln('Basic:     ',c1*sum+n1);
  if t>=sum then sek:=n2
  else sek:=n2+(sum-t)*c2;
  writeln('Combined:  ',sek);
  writeln('Unlimited: ',n3);
end.

please help me, why I've WA 1 ????

Edited by author 14.10.2007 22:49

Edited by author 14.10.2007 22:49