求双色球几连号及数量

最新推荐文章于 2025-06-24 11:31:43 发布

weixin_30339969

最新推荐文章于 2025-06-24 11:31:43 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/tzshowtime/p/4234402.html

功能很单一，如题。直接贴代码：

private void button1_Click(object sender, EventArgs e)
        {
            int[] iNumber = new int[6];
            iNumber[0] = 3;
            iNumber[1] = 4;
            iNumber[2] = 5;
            iNumber[3] = 15;
            iNumber[4] = 16;
            iNumber[5] = 18;

            
            Array.Sort(iNumber);

            int iCount2 = 0;
            int iCount3 = 0;
            int iCount4 = 0;
            int iCount5 = 0;
            int iCount6 = 0;

            for (int i = 0; i < 6; i++)
            {
                if (i + 1<6 && iNumber[i] + 1 == iNumber[i + 1])
                {
                    if (i + 2<6 && iNumber[i + 1] + 1 == iNumber[i + 2])
                    {
                        if (i + 3 < 6 && iNumber[i + 2] + 1 == iNumber[i + 3])
                        {
                            if (i + 4 < 6 && iNumber[i + 3] + 1 == iNumber[i + 4])
                            {
                                if (i + 5 < 6 && iNumber[i + 4] + 1 == iNumber[i + 5])
                                {
                                    iCount6 = iCount6 + 1;
                                    i = i + 5;
                                }
                                else
                                {
                                    iCount5 = iCount5 + 1;
                                    i = i + 4;
                                }
                            }
                            else
                            {
                                iCount4 = iCount4 + 1;
                                i = i + 3;
                            }
                        }
                        else
                        {
                            iCount3 = iCount3 + 1;
                            i = i + 2;
                        }
                    }
                    else
                    {
                        iCount2 = iCount2 + 1;
                        i = i + 1;
                    }
                }
            }
        }

转载于:https://www.cnblogs.com/tzshowtime/p/4234402.html