2021-03-12

最新推荐文章于 2025-05-14 10:28:08 发布

天涯思君不见君

最新推荐文章于 2025-05-14 10:28:08 发布

阅读量91

点赞数

文章标签：算法

本文链接：https://blog.youkuaiyun.com/m0_50426377/article/details/114703258

版权

HDU 1197(进制转换相关的问题)

题目链接

Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of its digits when represented in duodecimal (base 12) notation.

For example, the number 2991 has the sum of (decimal) digits 2+9+9+1 = 21. Since 2991 = 11728 + 8144 + 9*12 + 3, its duodecimal representation is 1893(12), and these digits also sum up to 21. But in hexadecimal 2991 is BAF16, and 11+10+15 = 36, so 2991 should be rejected by your program.

The next number (2992), however, has digits that sum to 22 in all three representations (including BB016), so 2992 should be on the listed output. (We don’t want decimal numbers with fewer than four digits - excluding leading zeroes - so that 2992 is the first correct answer.)

Input

There is no input for this problem.

Output

Your output is to be 2992 and all larger four-digit numbers that satisfy the requirements (in strictly increasing order), each on a separate line with no leading or trailing blanks, ending with a new-line character. There are to be no blank lines in the output. The first few lines of the output are shown below.

Sample Input

There is no input for this problem.

Sample Output

Code

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctype.h>
using namespace std;
int solove(char s[]);
int main()
{
    int num=2992;
    while(num<=9999)
    {
        int temp=num;
        int result1,result2,result3=0;//定义三个变量用来记录不同进制下的数字和
        char str[100];
        while (temp)    //result3用来记录num在十进制下的各个数字之和
        {
            result3+=(temp%10);
            temp/=10;
        }
        _itoa(num, str, 12);//itoa是一个库函数,包含在stdlib头文件中第一个参数是十进制的原始数据;把num转换成12进制的数字并且储存在str数组里面
        result1=solove(str);

        memset(str,0,sizeof(str));//把str数组清0,以便下次使用

        _itoa(num, str, 16);
        result2=solove(str);

        if (result3==result1&&result3==result2)
            printf("%d\n",num);

        num++;
    }
}
int solove(char str[])//这个函数主要是把存在str数组里面的那个转化后的值的每位数字的和
{
    int len=strlen(str);
    int result=0;
    for(int i=0;i<len;i++)
    {
        if (isalpha(str[i]))//判断是否是字母,也可以用ASII码来判断
            result=result+str[i]-'a'+10;//这里注意一下哟
        if (isdigit(str[i]))
            result=result+str[i]-'0';
    }

    return result;
}