• 2409 卒行走

    Time Limit : 6000/3000 MS(Java/Others) | Memory Limit : 65536/32768 KB(Java/Others)

    Submits : 404 | Solved : 113

    Description

    A点有一个卒,需要走到目标B点。行走规则:可以向下或者向右。要求计算从A能够到达B的路径的条数,并输出每一条路径。


    Input

    输入A点的坐标(ax,ay)和B点的坐标(bx,by),
    (|ax-bx|<=10且|ay-by|<=10).

    Output

    按照规定顺序输出路径(一行一条).
    最后输出路径数.
    如:
    (1,3)->(2,3)->(2,4)和(1,3)->(1,4)->(2,4)两条路径
    那么输出顺序为
    (1,3)->(1,4)->(2,4)
    (1,3)->(2,3)->(2,4)
    既先按x升序,再按y升序.
    具体见样例.

    Sample Input

    1 1 4 3

    Sample Output

    (1,1)-->(1,2)-->(1,3)-->(2,3)-->(3,3)-->(4,3)
    (1,1)-->(1,2)-->(2,2)-->(2,3)-->(3,3)-->(4,3)
    (1,1)-->(1,2)-->(2,2)-->(3,2)-->(3,3)-->(4,3)
    (1,1)-->(1,2)-->(2,2)-->(3,2)-->(4,2)-->(4,3)
    (1,1)-->(2,1)-->(2,2)-->(2,3)-->(3,3)-->(4,3)
    (1,1)-->(2,1)-->(2,2)-->(3,2)-->(3,3)-->(4,3)
    (1,1)-->(2,1)-->(2,2)-->(3,2)-->(4,2)-->(4,3)
    (1,1)-->(2,1)-->(3,1)-->(3,2)-->(3,3)-->(4,3)
    (1,1)-->(2,1)-->(3,1)-->(3,2)-->(4,2)-->(4,3)
    (1,1)-->(2,1)-->(3,1)-->(4,1)-->(4,2)-->(4,3)
    10

    HINT

    由于输出庞大

    请不要用cout

    请用printf


    Source

    NBU OJ

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