POJ 3641 Pseudoprime numbers 题解 《挑战程序设计竞赛》 � 码农场

POJ 3641 Pseudoprime numbers 题解 《挑战程序设计竞赛》 � 码农场

伪素数：满足①p不是素数②存在a > 1使得a^p = a (mod p)的p是伪素数，给出p和a，判断p是否是伪素数。

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Basic and Extended Euclidean algorithms - GeeksforGeeks

Basic and Extended Euclidean algorithms - GeeksforGeeks

Extended Euclidean Algorithm:
Extended Euclidean algorithm also finds integer coefficients x and y such that:

  ax + by = gcd(a, b) 

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How to efficiently sort a big list dates in 20's - GeeksforGeeks

How to efficiently sort a big list dates in 20's - GeeksforGeeks
A Simple Solution is to use a O(nLogn) algorithm like Merge Sort. We can sort the list in O(n) time using Radix Sort. In a typical Radix Sort implementation, we first sort by last digit, then by second last digit, and so on. Here we sort in following order.
1) First sort by day using counting sort
2) Then sort by month using counting sort
3) Finally sort by year using counting sort

void radixSortDates(Date arr[], int n)

{

   // First sort by day

   countSortDay(arr, n);

 

   // Then by month

   countSortMonth(arr, n);

 

   // Finally by year

   countSortYear(arr, n);

}

 

// A function to do counting sort of arr[] according to

// day

void countSortDay(Date arr[], int n)

{

    Date output[n]; // output array

    int i, count[31] = {0};

 

    // Store count of occurrences in count[]

    for (i=0; i<n; i++)

        count[arr[i].d - 1]++;

 

    // Change count[i] so that count[i] now contains

    // actual position of this day in output[]

    for (i=1; i<31; i++)

        count[i] += count[i-1];

 

    // Build the output array

    for (i=n-1; i>=0; i--)

    {

        output[count[arr[i].d - 1] - 1] = arr[i];

        count[arr[i].d - 1]--;

    }

 

    // Copy the output array to arr[], so that arr[] now

    // contains sorted numbers according to curent digit

    for (i=0; i<n; i++)

        arr[i] = output[i];

}

 

// A function to do counting sort of arr[] according to

// month.

void countSortMonth(Date arr[], int n)

{

    Date output[n]; // output array

    int i, count[12] = {0};

 

    for (i = 0; i < n; i++)

        count[arr[i].m - 1]++;

    for (i = 1; i < 12; i++)

        count[i] += count[i - 1];

    for (i=n-1; i>=0; i--)

    {

        output[count[arr[i].m - 1] - 1] = arr[i];

        count[arr[i].m - 1]--;

    }

    for (i = 0; i < n; i++)

        arr[i] = output[i];

}

 

// A function to do counting sort of arr[] according to

// year.

void countSortYear(Date arr[], int n)

{

    Date output[n]; // output array

    int i, count[1000] = {0};

    for (i = 0; i < n; i++)

        count[arr[i].y - 2000]++;

    for (i = 1; i < 1000; i++)

        count[i] += count[i - 1];

    for (i = n - 1; i >= 0; i--)

    {

        output[count[arr[i].y - 2000] - 1] = arr[i];

        count[arr[i].y - 2000]--;

    }

    for (i = 0; i < n; i++)

        arr[i] = output[i];

}

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