计算C中一个整数中特定数字的出现 [英] Counting occurance of a specific number in an integer in C

问题描述

我遇到了一些麻烦...我试图输入一个数字，然后输入多个数字的整数.我试图计算第一个数字在整数中出现了多少次.现在，我编写了这个非常简单的代码，向您展示了我实际上正在尝试做的事情.关键是，此代码仅比较两个整数，并告诉我它们是否相同.请注意，我对C编程非常缺乏经验，因此这个问题...

I'm having some troubles... I'm trying to input a number, followed by an integer of multiple numbers. I am trying to count how many times the first number occurs in the integer. Now, I made this really easy code to show you what I actually am trying to do. The thing is, this code only compares two integers and tells me if they are the same or not. Mind you, I am very unexperienced in C programming, hence this question...

int main(){

    int numberOne;
    int numberTwo;
    int count = 0;

    scanf("%d", &numberOne);
    scanf("%d", &numberTwo);

    if(numberOne == numberTwo){
        count++;
    }

    printf("Amount of equals found: %d", count);

return 0;
}

现在，如果我有以下输入:'1 1021023234'，则输出将是:'发现的等于数量:0'输出(在这种情况下)应为输出将为等于找到的数量:2"

Now, if I'd have the following input: '1 1021023234', the output would be: 'Amount of equals found:0' The output should be (in this case) 'The output would be Amount of equals found:2'

我希望你们能给我一些提示.

I hope you guys can give me some tips.

推荐答案

如果我正确理解，则需要以下内容

If I have understood correctly then you need the following

#include <stdio.h>

int main(void) 
{
    unsigned int numberOne;
    unsigned int numberTwo;
    unsigned int x;
    unsigned int base;
    size_t n;

    printf( "Enter first number: " );
    scanf( "%u", &numberOne );

    printf( "Enter second number: " );
    scanf( "%u", &numberTwo );

    x = numberOne;
    base = 1;
    do { base *= 10; } while ( x /= 10 );

    n = 0;

    do { n += numberOne == numberTwo % base; } while ( numberTwo /= 10 );

    printf( "Amount of equals found: %u", n );

    return 0;
}

对于数字 12 和 76512612 ，输出为

Amount of equals found: 2

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