CSES - Datatähti 2023 loppu - Results
Submission details
Task:Merkkijonot
Sender:cppbetter
Submission time:2023-11-26 21:36:17 +0200
Language:C++20
Status:COMPILE ERROR

Compiler report

input/code.cpp:1:1: error: '\U00003005' does not name a type
    1 | 々
      | ^~
In file included from /usr/include/c++/11/iosfwd:40,
                 from /usr/include/c++/11/ios:38,
                 from /usr/include/c++/11/ostream:38,
                 from /usr/include/c++/11/iostream:39,
                 from input/code.cpp:2:
/usr/include/c++/11/bits/postypes.h:98:11: error: 'ptrdiff_t' does not name a type
   98 |   typedef ptrdiff_t     streamsize; // Signed integral type
      |           ^~~~~~~~~
/usr/include/c++/11/bits/postypes.h:41:1: note: 'ptrdiff_t' is defined in header '<cstddef>'; did you forget to '#include <cstddef>'?
   40 | #include <cwchar> // For mbstate_t
  +++ |+#include <cstddef>
   41 | 
In file included from /usr/include/c++/11/bits/exception_ptr.h:40,
                 from /usr/include/c++/11/exception:153,
                 from /usr/include/c++/11/ios:39,
                 from /usr/include/c++/11/ostream:38,
                 from /usr/include/c++/11/io...

Code

々
#include <iostream>
#include <string>
#include <vector>
#include <tuple>
#include <unordered_map>
 
std::vector<std::pair<int, int>> pairs;
int mem[500*500*500];
 
// This is O(2^n) algo
int check(int i, std::pair<int, int> val)
{
    if(val.first == 0 && val.second == 0)
        return 1;
 
    if(i == pairs.size())
        return 0;
 
    auto a = mem[i*500*500+val.first*500 +val.second];
    if(a != -1)
        return a;
 
    std::pair<int, int> test;
    test.first = val.first - pairs[i].first;
    test.second = val.second - pairs[i].second;
    
    int sum = 0;
    if(test.second >= 0 && test.first >= 0)
        sum += check(i + 1, test) % 1000000007;
    sum += check(i + 1, val) % 1000000007;
 
    mem[i*500*500+val.first*500 +val.second] = sum;
    return sum % 1000000007;
}
 
int main()
{
    int n;
    std::cin >> n;
 
    // Sub sum problem except sketchy???? so best case O(nm) ?????
    // Only half has to be cheked
    // pari (count of a, count of b) for 1 part
 
    pairs.resize(n);
    int totalA = 0;
    int totalB = 0;
 
    for(int i = 0; i < n; i++)
    {
        std::string str;
        std::cin >> str;
 
        // Count the intsances of a
        for(auto c : str)
        {
            if(c == 'a')
            {
                totalA++;
                pairs[i].first++;
            }
            else {
                totalB++;
                pairs[i].second++;
            }
        }
    }
 
    for(auto& a : mem)
        a = -1;
 
    //mem = { -1 };
    if(totalA % 2 == 0 && totalB % 2 == 0)
        std::cout << check(0, { totalA/2, totalB/2 }) << "\n";
    else
        std::cout << 0 << "\n";
}