[Problem 17]欧拉

这一题乍看上去比较困难。稍作思考就会发现其算法非常明显。具体思路下面的代码的comment中写得很清楚了。（第一遍实现后发现post到欧拉上说答案不对；后来发现是11X， 21X和12X,24X的计算方法不同这点没有考虑到。更正算法，终于得到正确答案）

#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    // 1 - 1000 word number calculation.
    //
    unsigned long sum = 0;

    // 1-19. Rule: as every word in this range is special so we have to write and calculate every word.           
    unsigned long arrA[] = { 
        0,
        3/*one*/, 
        3/*two*/, 
        5/*three*/, 
        4/*four*/,
        4/*five*/,
        3/*six*/,
        5/*seven*/,
        5/*eight*/,
        4/*nine*/,
        3/*ten*/,
        6/*eleven*/,
        6/*twelve*/,
        8/*thirteen*/,
        8/*fourteen*/,
        7/*fifteen*/,
        7/*sixteen*/,
        9/*seventeen*/,
        8/*eighteen*/,
        8/*nineteen*/
    };

    for(int i = 1; i < 20; i++)
    {
        sum += arrA[i];
    }

    // 20-99. Rule: for example, 24-- twenty four. Letters it uses == number of (twenty) arrB[] + number of (four) arrA[].
    unsigned long arrB[] = { 
        0,
        3/*"ten"*/, 
        6/*"twenty"*/, 
        6/*"thirty"*/, 
        5/*"forty"*/,  
        5/*"fifty"*/, 
        5/*"sixty"*/, 
        7/*"seventy"*/, 
        6/*"eighty"*/, 
        6/*"ninety"*/};
    int geWei;
    int shiWei;
    for (int i = 20; i <= 99; i++ )
    {
        geWei = i % 10;
        if(geWei > 0)
            sum += arrA[geWei];

        shiWei = i / 10;
        if(shiWei > 0)
            sum += arrB[shiWei];
    }

    // 100-999. The rule is: for example, 112 == (one) (hundred and) (twelve); 123 == (one) (hundred and) (twenty) (three). You can
    // see clearly where to retrieve each part's letter number.
    unsigned long nNumA = 7;//"hundred";
    unsigned long nNumB = 10; // "hundred and"
    int baiWei = 0; // 123's baiWei is 1.
    int yushu = 0; // 123's yushu is 23.
    bool bNotSingle = false; // 100 is single while 101 is notSingle.
    for (int i = 100; i <= 999; i++)
    {
        bNotSingle = true;

        yushu = i % 100;
        if(yushu > 0 && yushu < 20)
            sum += arrA[yushu];
        else if(yushu >= 20)
        {
            geWei = yushu % 10;
            if(geWei > 0)
                sum += arrA[geWei];

            shiWei = yushu / 10;
            if(shiWei > 0)
                sum += arrB[shiWei];
        }
        else
            bNotSingle = false;        

        baiWei = i / 100;
        if (bNotSingle)
            sum = sum + nNumB + arrA[baiWei];
        else
            sum = sum + nNumA + arrA[baiWei];        

    }

    // 1000
    sum += 11; // "one thousand".

    cout << "sum:" << endl;
    cout << sum << endl;           


    return 0;
}