• USACO Training 1.5 Arithmetic Progressions


    An arithmetic progression is a sequence of the form a, a+b, a+2b, ..., a+nb where n=0, 1, 2, 3, ... . For this problem, a is a non-negative integer and b is a positive integer.

    Write a program that finds all arithmetic progressions of length n in the set S of bisquares. The set of bisquares is defined as the set of all integers of the form p2 + q2 (where p and q are non-negative integers).

    TIME LIMIT: 5 secs

    PROGRAM NAME: ariprog

    INPUT FORMAT

    Line 1:N (3 <= N <= 25), the length of progressions for which to search
    Line 2:M (1 <= M <= 250), an upper bound to limit the search to the bisquares with 0 <= p,q <= M.

    SAMPLE INPUT (file ariprog.in)

    5
    7
    

    OUTPUT FORMAT

    If no sequence is found, a single line reading `NONE'. Otherwise, output one or more lines, each with two integers: the first element in a found sequence and the difference between consecutive elements in the same sequence. The lines should be ordered with smallest-difference sequences first and smallest starting number within those sequences first.

    There will be no more than 10,000 sequences.

    SAMPLE OUTPUT (file ariprog.out)

    1 4
    37 4
    2 8
    29 8
    1 12
    5 12
    13 12
    17 12
    5 20
    2 24
    1. /*
    2. ID: choiyin1
    3. PROG: ariprog
    4. LANG: C++
    5. */
    6. #include
    7. using namespace std;
    8. int main() {
    9. freopen("ariprog.in", "r", stdin);
    10. freopen("ariprog.out", "w", stdout);
    11. int n;
    12. int m;
    13. bool pq[127000] = {};
    14. cin >> n >> m;
    15. int squareNum = 0;
    16. for (int i = 0; i <= m; i ++){
    17. for (int j = i; j <= m; j ++){
    18. pq[i*i + j*j] = true;
    19. }
    20. }
    21. int step;
    22. int start;
    23. int maxInPq = m*m + m*m;
    24. bool b = false;
    25. for (step = 1; step <= maxInPq; step ++){
    26. for (start = 0; start <= maxInPq; start ++){
    27. if (start + (n-1) * step > maxInPq)
    28. break;
    29. if (!pq[start])
    30. continue;
    31. int i, pos = start;
    32. for (i = 1; i < n && pq[pos]; i ++){
    33. pos += step;
    34. }
    35. if (i == n && pq[pos]){
    36. printf("%d %d\n", start, step);
    37. //cout << start << endl;
    38. b = true;
    39. }
    40. }
    41. }
    42. if (!b)
    43. cout << "NONE" << endl;
    44. return 0;
    45. }

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  • 原文地址:https://blog.csdn.net/GeekAlice/article/details/127097112