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docs: fix grammatical errors and typos (#2201)
* docs: fix grammatical errors and typos
* compilation error fixed
* Revert "compilation error fixed"
This reverts commit 0083cbfd1a
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@ -39,13 +39,13 @@ namespace dynamic_programming {
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namespace knapsack {
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/**
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* @brief Picking up all those items whose combined weight is below
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* given capacity and calculating value of those picked items.Trying all
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* the given capacity and calculating the value of those picked items. Trying all
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* possible combinations will yield the maximum knapsack value.
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* @tparam n size of the weight and value array
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* @param capacity capacity of the carrying bag
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* @param weight array representing weight of items
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* @param value array representing value of items
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* @return maximum value obtainable with given capacity.
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* @param weight array representing the weight of items
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* @param value array representing the value of items
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* @return maximum value obtainable with a given capacity.
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*/
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template <size_t n>
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int maxKnapsackValue(const int capacity, const std::array<int, n> &weight,
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@ -53,7 +53,7 @@ int maxKnapsackValue(const int capacity, const std::array<int, n> &weight,
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std::vector<std::vector<int> > maxValue(n + 1,
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std::vector<int>(capacity + 1, 0));
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// outer loop will select no of items allowed
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// inner loop will select capcity of knapsack bag
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// inner loop will select the capacity of the knapsack bag
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int items = sizeof(weight) / sizeof(weight[0]);
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for (size_t i = 0; i < items + 1; ++i) {
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for (size_t j = 0; j < capacity + 1; ++j) {
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@ -62,22 +62,22 @@ int maxKnapsackValue(const int capacity, const std::array<int, n> &weight,
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// will be zero
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maxValue[i][j] = 0;
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} else if (weight[i - 1] <= j) {
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// if the ith item's weight(in actual array it will be at i-1)
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// if the ith item's weight(in the actual array it will be at i-1)
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// is less than or equal to the allowed weight i.e. j then we
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// can pick that item for our knapsack. maxValue will be the
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// obtained either by picking the current item or by not picking
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// current item
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// picking current item
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// picking the current item
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int profit1 = value[i - 1] + maxValue[i - 1][j - weight[i - 1]];
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// not picking current item
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// not picking the current item
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int profit2 = maxValue[i - 1][j];
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maxValue[i][j] = std::max(profit1, profit2);
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} else {
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// as weight of current item is greater than allowed weight, so
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// maxProfit will be profit obtained by excluding current item.
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// as the weight of the current item is greater than the allowed weight, so
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// maxProfit will be profit obtained by excluding the current item.
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maxValue[i][j] = maxValue[i - 1][j];
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}
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}
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@ -90,7 +90,7 @@ int maxKnapsackValue(const int capacity, const std::array<int, n> &weight,
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} // namespace dynamic_programming
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/**
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* @brief Function to test above algorithm
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* @brief Function to test the above algorithm
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* @returns void
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*/
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static void test() {
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@ -18,8 +18,8 @@ int main() {
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count++;
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}
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/* Calaculation for checking of armstrongs number i.e.
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in a n digit number sum of the digits raised to a power of n
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/* Calculation for checking of armstrongs number i.e.
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in an n-digit number sum of the digits is raised to a power of n
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is equal to the original number */
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temp = n;
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