自己备查：如何优化算法--缓存

 题目:

 

http://projecteuler.net/index.php?section=problems&id=164

 

projecteuler.net #164

 

 

 

How many 20 digit numbers n (without any leading zero) exist such that no three consecutive digits of n have a sum greater than 9?

 

使用简单算法，从高位到低位便利每种组合，慢的要死

 

        List<int> number = new List<int>();
        int count = 0;
        public object Run()
        {

            for (int i = 0; i < 10; i++)
            {
                //number.Add(i);
                Run2(1, i, 0);
                //number.RemoveAt(0);
            }

            return count;
        }

        void Run2(int level, int last, int lastlast)
        {
            for (int i = 0, length = 9 - last - lastlast; i <= length; i++)
            {
                if (level == 2)
                {
                    Console.WriteLine("" + lastlast + last + i);
                }
                //if (level > 1 && last + lastlast + i > 9)
                //    return;

                if (level == 6)
                {
                    count += 1;
                    return;
                }

                //number.Add(i);
                Run2(level + 1, i, last);
                //number.RemoveAt(level);
            }
        }

 

优化后，从低位到高位递增，每次缓存当位所有ok的数字以及对应的可能性供下位处理，时间复杂度O(n),瞬间完成

    class Problem164 : IProblem
    {
        Dictionary<int, C> last = new Dictionary<int, C>();
        Dictionary<int, C> current = new Dictionary<int, C>();
        public object Run()
        {
            int l, ll;
            for (int i = 0; i < 100; i++)
            {
                l = i / 10;
                ll = i % 10;
                if (l + ll < 10)
                    last.Add(i, new C(l, ll, 1));
            }
            return Run2(2);
        }

        public long Run2(int level)
        {
            int thisLevelBegin = (level == 19 ? 1 : 0);

            current = new Dictionary<int, C>();

            int l, ll, sum, newNum;
            long count;
            foreach (KeyValuePair<int, C> p in last)
            {
                l = p.Value.L;
                ll = p.Value.LL;
                count = p.Value.count;
                sum = p.Value.sum;
                for (int i = thisLevelBegin, length = 10 - sum; i < length; i++)
                {
                    newNum = i * 10 + l;
                    if (!current.ContainsKey(newNum))
                    {
                        current.Add(newNum, new C(i, l, count));
                    }
                    else
                    {
                        current[newNum].count += count;
                    }
                }
            }
            if (level == 19)
            {
                count = 0;
                foreach (KeyValuePair<int, C> p in current)
                {
                    count += p.Value.count;
                }

                return count;
            }
            last = current;
            return Run2(level + 1);
        }
    }

    class C
    {
        public int LL;
        public int L;
        public int sum;
        public long count;
        public C(int l, int ll, long c)
        {
            LL = ll;
            L = l;
            count = c;
            sum = l + ll;
        }
    }