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  • 十进制数1~n中1出现的次数

    //copyright@ saturnman
    //updated@ 2011 July
    #include "stdafx.h"
    #include "string.h"
    #include "stdlib.h"

    int NumberOf1(const char* strN);
    int PowerBase10(unsigned int n);

    /////////////////////////////////////////////////////////////////////////////
    // Find the number of 1 in an integer with radix 10
    // Input: n - an integer
    // Output: the number of 1 in n with radix
    /////////////////////////////////////////////////////////////////////////////
    int NumberOf1BeforeBetween1AndN_Solution2(int n)
    {
    if(n <= 0)
    return 0;

    // convert the integer into a string
    char strN[50];
    sprintf(strN, "%d", n);

    return NumberOf1(strN);
    }


    /////////////////////////////////////////////////////////////////////////////
    // Find the number of 1 in an integer with radix 10
    // Input: strN - a string, which represents an integer
    // Output: the number of 1 in n with radix
    /////////////////////////////////////////////////////////////////////////////
    int NumberOf1(const char* strN)
    {
    if(!strN || *strN < '0' || *strN > '9' || *strN == '\0')
    return 0;

    int firstDigit = *strN - '0';
    unsigned int length = static_cast<unsigned int>(strlen(strN));

    // the integer contains only one digit
    if(length == 1 && firstDigit == 0)
    return 0;

    if(length == 1 && firstDigit > 0)
    return 1;

    // suppose the integer is 21345
    // numFirstDigit is the number of 1 of 10000-19999 due to the first digit
    int numFirstDigit = 0;
    // numOtherDigits is the number of 1 01346-21345 due to all digits
    // except the first one
    int numOtherDigits = firstDigit * (length - 1) * PowerBase10(length - 2);
    // numRecursive is the number of 1 of integer 1345
    int numRecursive = NumberOf1(strN + 1);

    // if the first digit is greater than 1, suppose in integer 21345
    // number of 1 due to the first digit is 10^4. It's 10000-19999
    if(firstDigit > 1)
    numFirstDigit = PowerBase10(length - 1);

    // if the first digit equals to 1, suppose in integer 12345
    // number of 1 due to the first digit is 2346. It's 10000-12345
    else if(firstDigit == 1)
    numFirstDigit = atoi(strN + 1) + 1;

    return numFirstDigit + numOtherDigits + numRecursive;
    }

    /////////////////////////////////////////////////////////////////////////////
    // Calculate 10^n
    /////////////////////////////////////////////////////////////////////////////
    int PowerBase10(unsigned int n)
    {
    int result = 1;
    for(unsigned int i = 0; i < n; ++ i)
    result *= 10;

    return result;
    }
    int num1Ofn(int n)
    {
    int count=0;
    while(n>0)
    {
    if(n%10 == 1)
    count++;
    n/=10;
    }
    return count;
    }
    int count1Num(int n)
    {
    int sum=0;
    for (int i =1;i <= n;i ++)
    {
    sum+=num1Ofn(i);
    }
    return sum;
    }
    int main()
    {
    printf("%d\n", count1Num(21345));
    printf("%d\n", NumberOf1BeforeBetween1AndN_Solution2(21345));
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/mfryf/p/2670404.html
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