笨拙的手指

• 方法一
• 方法二

方法一

用暴力枚举法枚举出二进制所有的情况，化为十进制数存放在一个数组中，
然后再去枚举三进制所有的情况，化为十进制一一去和数组中的元素去匹配，如果相等就说明找到了答案，结束程序

1.将m进制的数转为十进制

int get(char* a, int m)
{
    int ret = 0;
    int len = strlen(a);
    for (int i = 0; i < len; i++)
    {
        ret = ret * m + (a[i]-'0');
    }
    return ret;
}

2.去数组中找相等的数

int find(int* arr, int a,int len1)
{

    for (int i = 0; i < len1; i++)
    {
        if (a == arr[i])
            return 1;

    }
    return 0;
}

3.完整代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int get(char* a, int m)
{
    int ret = 0;
    int len = strlen(a);
    for (int i = 0; i < len; i++)
    {
        ret = ret * m + (a[i]-'0');
    }
    return ret;
}
int find(int* arr, int a,int len1)
{

    for (int i = 0; i < len1; i++)
    {
        if (a == arr[i])
            return 1;

    }
    return 0;
}
int main()
{

    char a[11], b[11];
    scanf("%s", a);
    scanf("%s", b);
    int len1 = strlen(a);
    int len2 = strlen(b);
    int* ret1 = (int*)malloc(sizeof(int) * len1);
    for (int i = 0; i < len1; i++)
    {
        a[i] ^= 1;
        ret1[i] = get(a, 2);
        a[i] ^= 1;
    }
    int p = 0;
    for (int i = 0; i < len2; i++)
    {
        char t = b[i];
        for (int j = 0; j < 3; j++)
        {
            b[i] = j + '0';
            int c = get(b, 3);
            if (find(ret1, c,len1) == 1)
            {
                printf("%d",c);
                return 0;
            }
            b[i] = t;
        }
    }
    return 0;
}

方法二

枚举二进制的所有情况，化为三进制后，去和b比较，假如仅有一位不相同，则说明该情况就是正确答案，程序结束

1. 二进制转化为十进制

int get(char* a, int m)
{
    int ret = 0;
    int len = strlen(a);
    for (int i = 0; i < len; i++)
    {
        ret = ret * m + (a[i] - '0');
    }
    return ret;
}

2.二进制转化为三进制

char* cp(char* a, int m)
{
    char* arr = (char*)malloc(sizeof(char) * 10);
    int temp = get(a, m);
    int i = 0;
    while (temp)
    {
        arr[i] = temp % 3 + '0';
        temp /= 3;
        i++;
    }
    arr[i] = '\0';
    for (int j = 0; j < i / 2; j++)
    {
        char t = arr[j];
        arr[j] = arr[i - j - 1];
        arr[i - j - 1] = t;
    }
    return arr;
}

3.统计每位不一样的个数

int count(char* a, char* b)
{
    int len1 = strlen(a);
    int len2 = strlen(b);
    if (len1 != len2)
        return 0;
    int count = 0;
    while (len2 >= 0 && len1 >= 0)
    {
        if (a[len1] != b[len2])
            count++;
        len1--;
        len2--;
    }
    return count;
}

4.完整代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
//二转十
int get(char* a, int m)
{
    int ret = 0;
    int len = strlen(a);
    for (int i = 0; i < len; i++)
    {
        ret = ret * m + (a[i] - '0');
    }
    return ret;
}
//二转三
char* cp(char* a, int m)
{
    char* arr = (char*)malloc(sizeof(char) * 10);
    int temp = get(a, m);
    int i = 0;
    while (temp)
    {
        arr[i] = temp % 3 + '0';
        temp /= 3;
        i++;
    }
    arr[i] = '\0';
    for (int j = 0; j < i / 2; j++)
    {
        char t = arr[j];
        arr[j] = arr[i - j - 1];
        arr[i - j - 1] = t;
    }
    return arr;
}
int count(char* a, char* b)
{
    int len1 = strlen(a);
    int len2 = strlen(b);
    if (len1 != len2)
        return 0;
    int count = 0;
    while (len2 >= 0 && len1 >= 0)
    {
        if (a[len1] != b[len2])
            count++;
        len1--;
        len2--;
    }
    return count;
}
int main()
{
    char a[30], b[30];
    scanf("%s", a);
    scanf("%s", b);
    int len1 = strlen(a);
    int len2 = strlen(b);
    int* ret1 = (int*)malloc(sizeof(int) * len1);
    for (int i = 0; i < len1; i++)
    {
        a[i] ^= 1;
        int t = count(b, cp(a, 2));
        if (t == 1)
        {
            printf("%d", get(a, 2));
            return 0;
        }
        a[i] ^= 1;
    }
}