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Commit 675e3b1

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3 files changed

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Lines changed: 43 additions & 0 deletions
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The beginning element is either A[i] - 1, A[i], A[i] + 1
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Let us notice that the first element uniquely determines the whole sequence.
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We will check if any of these 3 options leads to a consecutive segment.
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-----
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int is_possible(vector <int> &A, int n)
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{
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for(int i = 1, current = n; i < A.size(); i++, current++)
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{
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if(abs(A[i] - current) > 1)
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{
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return false;
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}
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}
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return true;
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}
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void solve()
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{
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int no_of_elements;
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cin >> no_of_elements;
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vector <int> A(no_of_elements + 1);
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for(int i = 1; i <= no_of_elements; i++)
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{
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cin >> A[i];
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}
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int possible = false;
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for(int beginning = A[1] - 1; beginning <= A[1] + 1; beginning++)
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{
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if(is_possible(A, beginning))
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{
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possible = true;
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}
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}
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cout << (possible ? "Yes" : "No") << "\n";
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}
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Let us keep an array which will tell us the maximum number of times we can buy a prefix of length i
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Let P[i] be the perfix sum of A[1, ... i]
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We can buy the prefix for i days if
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P[i] + (x - 1)i <= Budget
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We can binary search for the value of x for each array element.
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------
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After that, we will try to find out the number of times we have bought a segment ending at i, exactly
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In order to do this, we will notice that we can buy the prefix ending at i
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no_of_days[i] - no_of_days[i + 1] times.
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-----
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void solve()
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{
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int no_of_elements, budget;
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cin >> no_of_elements >> budget;
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vector <int> A(no_of_elements + 1);
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for(int i = 1; i <= no_of_elements; i++)
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{
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cin >> A[i];
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}
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sort(A.begin(), A.end());
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vector <long long> prefix_sum(no_of_elements + 1);
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for(int i = 1; i <= no_of_elements; i++)
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{
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prefix_sum[i] = prefix_sum[i - 1] + A[i];
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}
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vector <int> no_of_days_used(no_of_elements + 2, 0);
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for(int i = 1; i <= no_of_elements; i++)
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{
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if(prefix_sum[i] > budget)
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{
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break;
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}
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long long low = 1, high = 1e9 + 1;
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while(high - low > 1)
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{
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long long mid = (low + high)/2;
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if(( prefix_sum[i] + (mid - 1)*i ) > budget)
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{
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high = mid;
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}
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else
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{
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low = mid;
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}
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}
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no_of_days_used[i] = low;
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//cout << "i = " << i << " Days = " << no_of_days_used[i] << "\n";
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}
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long long answer = 0;
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for(int i = 1; i <= no_of_elements; i++)
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{
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answer += (no_of_days_used[i] - no_of_days_used[i + 1])*i;
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}
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cout << answer << "\n";
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}
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If the length of the segment is even, we can just keep using blocks of 2
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If the length of the segment is odd, we can use one blocks of 3, after which it is even and reduced to the case above.
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The only time it is impossible is when the length of the segment is 1.
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-----
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void solve()
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{
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string S;
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cin >> S;
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int possible = (S.size() == 1 ? false : true);
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for(int i = 0; i < S.size(); i++)
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{
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if(i == 0)
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{
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if(S[i] != S[i + 1])
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{
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possible = false;
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}
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continue;
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}
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if(i + 1 == S.size())
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{
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if(S[i] != S[i - 1])
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{
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possible = false;
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}
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break;
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}
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if(S[i - 1] != S[i] && S[i] != S[i + 1])
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{
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possible = false;
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break;
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}
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}
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cout << (possible ? "YES" : "NO") << "\n";
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}

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