Question

我的代码问题我需要制作。我必须从命令行中获取14个参数并使用它们来制作彩票号码，中奖号码，然后将这些参数进行比较。

例如，使用此参数：./a.out 2 30 17 8 6 19 24 7 6 1 2 3 5 4

应该做这样的事情：

Winning numbers: 2 30 17 8 6 19 24
Lottonumbers: 7 6 1 2 3 5 4

2 are the same: 6 2

我的代码几乎按预期工作，但我似乎无法正确打印： 2是相同的。它总是像这样循环：1 are the same: 6 2 are the same: 2。

数字2是比较2个数组时找到的相同数字的数量。我的问题是如何打印它以便它不会复制文本和适当数量？即使它如此简单，我的头似乎也无法工作：/

#include <stdio.h>
#include <stdlib.h>

int main(int args, char **argv)
{    
    int i;
    int winningNumbers[7];
    int lottoNumbers[7];
    int j;
    int a;
    int b;
    int winningNumber;
    int lottoNumber;
    int count = 0;

    printf("Winning numbers: ");

    for (i=0;i<7; i++) {    
        winningNumber = atoi(argv[i+1]);    
        winningNumbers[i] = winningNumber;    
        printf("%d ", winningNumber);    
    }

    printf("\n");    
    printf("Lotto numbers:: ");

    for (j= 8; j < args; j++) {

        lottoNumber = atoi(argv[j]);    
        lottoNumbers[j-8] = lottoNumber;    
        printf("%d ", lottoNumber);

    }
    printf("\n");

    for(a = 0; a < 7; a++) {

        for(b=0; b < 7; b++) {
            if (lottoNumbers[a] == winningNumbers[b]) {
                count = count + 1;
                printf("%d are the same: %d", count, winningNumbers[b]);

            }
        }
    }

    return 0;
}

Answer 1

搜索匹配项并显示结果是两项独立的任务。不尝试同时进行操作更简单，更灵活。

首先搜索匹配并将它们存储在数组中。然后根据需要显示数组的内容。

int main (int argc, char *argv[])
{
    int winningNumbers[7];
    int lottoNumbers[7];
    int commonNumbers[7];
    int count = 0;

    // fill winningNumbers
    // fill lottoNumbers

    // NOTE: the following loop assumes that in both arrays
    // no number is repeated.
    // You should check that this is indeed the case.
    for (int i = 0; i < 7; i++) {
        for (int j = 0; j < 7; j++) {
            if (lottoNumbers[i] == winningNumbers[j]) {
                commonNumbers[count] = lottoNumbers[i];
                count++;
            }
        }
    }

    printf ("%d are the same:", count);
    for (int i = 0; i < count; i++) {
        printf (" %d", commonNumbers[i]);
    }
    printf ("\n");

    return 0;
}

许多简单的程序应遵循以下结构：

阅读并检查输入
将输入转换为输出
打印输出

Answer 2

for(b=0; b < 7; b++) {
    if (lottoNumbers[a] == winningNumbers[b]) {
        count = count + 1;
    }
}
printf("%d are the same: ", count);
for(b=0; b < 7; b++) {
    if (lottoNumbers[a] == winningNumbers[b]) {
    printf(" %d", winningNumbers[b]);
    }
}
printf("\n");

Answer 3

int finalArray[7];
int i;
for(a = 0; a < 7; a++) {
    for(b=0; b < 7; b++) {
        if (lottoNumbers[a] == winningNumbers[b]) {
            finalArray[count] = lottoNumbers[a];
            count = count + 1;
        }
    }
}
printf("%d are same: ", count);
for(i = 0; i < count; i++)
    printf("%d ", finalArray[i]);

如何打印我的2个数组有相同整数的数量？ [C]

3 个答案: