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Problem solution · C++

CCC 2017 S1 - Sum Game

CCC 2017 S1 - Sum Game: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Dynamic programming
Source
CCCSolutions
Length
40 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For CCC 2017 S1 - Sum Game, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 40 lines of C++ from the credited upstream file ccc17s1.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 2 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2017 S1 - Sum Game · C++C++
Use this to learn the idea, then write your own version.
// CCC 2017 Senior 1: Sum Game // by John Liao, Lo-Ellen Park S.S. // This problem asks to keep track of the last time the two teams have the same sum. // We use totalSum to record the difference between the sums of the two teams. // When totalSum reaches 0, we know that the two teams have the same sum.  // Runtime: O(N) complexity// Memory: O(N) complexity #include<iostream>#include <vector> using namespace std; int main() {    int N, totalSum = 0, lastequal = 0;     cin >> N;     vector<int> A(N);    for (int i = 0; i < N; i++) {        cin >> A[i];    }      for (int i = 0; i < N; i++) {        int b; cin >> b;         totalSum += A[i] - b;         if (totalSum == 0) lastequal = i + 1;    }     cout << lastequal << endl;     return 0;}

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