count.txt

统计数字问题： 一本书的页码从自然数1开始顺序编码直到自然数n。 书的页码按照通常的习惯编排

问题描述：

    一本书的页码从自然数1开始顺序编码直到自然数n。 书的页码按照通常的习惯编排，
每个页码都不含多余的前导数字0。例如第6页用数字6表示，而不是006，06等。数字技术问
题要求对于给定的书的总页码n，计算出书的全部页码中分别用到多少数字0，1，2，3，4，
5，6，7，8?
// Count the digital in the number from 0 to n.
void CountFromLeft( int n = 0 )
{
    unsigned long countArray[10] = {0};

    // 10 for integer, one fore '\0'
    char maxNumber[11] = {0};
    if( n > 0 )
    {
        sprintf( maxNumber, "%d", n );
        int cursor = 0;
        int numberCount = 0;
        while( maxNumber[ cursor ] != '\0' )
        {
            // Add one digital at the end, the count of the
                           // number will be about 10 double. So the count of

                           // left digital will be approximately 10 times.
            for( int i=0; i<10; i++ )
            {
                countArray[i] *= 10;
            }
            // But not all
            // For 2=>23, '1' 10 double, '2' only 3+1,
            // For 2=>29, '2' will 10 dobule too.
            for( int j=0; j<cursor; j++ )
            {
                countArray[ maxNumber[ j ] - '0' ]
                                     -= '9' - maxNumber[ cursor ];
            }

            // There are some new digital for the last digit.
            // Add one for each digital.
            for( int i=0; i<10; i++ )
            {
                countArray[i] += numberCount;
            }
            // Not all again.
            for( int i=1; i<=( maxNumber[ cursor ] - '0' ); i++)
            {
                countArray[i]++;
            }

            // Get the count for previous part.
            numberCount *= 10;
            numberCount += maxNumber[ cursor ] - '0';

            cursor++;
        }
    }

    // Output the result
    cout << "Each digital:\n";
    int Total = 0;
    for( int i=0; i<10; i++ )
    {
        cout << "\t[" << i << "]: " << countArray[i] << "\n";
        Total += countArray[i];
    }
    cout << "Total: " << Total << "\n";
}
从右到左的

这里的思路是如果你拿到的参数是103
你先算从1到3有多少个数字
再算03,因为03和3是一样的,所以这个地方就不加了
然后算03到103增加了多少

// Count the digital in the number from 0 to n.
void CountFromRight( int n = 0 )
{
    unsigned long countArray[10] = {0};
    unsigned int power[10] = {1,10,100,1000,10000,100000,1000000,10000000,1000
00000,1000000000};

    int currentValue = n;
    int currentPos = 0;
    int addedNumber = 0;
    // Initialize here only for security, you can remove next two lines.
    int nextValue = 0;
    int currentDigital = 0;
    while( currentValue > 0 )
    {
        nextValue = currentValue / 10;
        currentDigital = currentValue - nextValue * 10;

        int currentPower = 0;
        // Add the number of every digital( 1/10 of added number by this digit
 )
        if( currentPos > 0 )
        {
            currentPower = currentDigital * currentPos * power[ currentPos - 1
 ];
            for( int j=0; j<10; j++)
            {
                countArray[ j ] += currentPower;
            }
            // not all
            if( addedNumber < power[ currentPos - 1 ] - 1 )
            {
                countArray[ 0 ] -= power[ currentPos - 1 ] - 1 - addedNumber;

            }
        }

        // Add the number of the less digital at top digit.
        currentPower = power[ currentPos ];
        for( int j=1; j<currentDigital; j++)
        {
            countArray[ j ] += currentPower;
        }

        // Add the number of current digital at top digit.
        // You can move the last for into this if
        if( currentDigital > 0 )
        {
            countArray[ currentDigital ] += addedNumber + 1;
        }


        currentValue = nextValue;
        addedNumber += currentDigital * power[ currentPos ];
        currentPos++;
    }


    // Output the result
    cout << "Each digital:\n";
    int Total = 0;
    for( int i=0; i<10; i++ )
    {
        cout << "\t[" << i << "]: " << countArray[i] << "\n";
        Total += countArray[i];
    }
    cout << "Total: " << Total << "\n";
}