1023. Have Fun with Numbers (20)-蒲公英云

1023. Have Fun with Numbers (20)

清疚 2022-05-31 10:05 275阅读 0赞

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

题目大意：

注意到数字123456789是一个9位数，有数字1到数字9组成，没有重复。我们将得到246913578，这恰好是另一个9为数字，只是排列顺序不同。如果我们再把他翻倍，看看结果如何！
现在您的任务是检查是否有更多的数字由此属性。也就是说给定一个K位数字加倍，你的任务是判断结果数字是否由原来的数字组成。
输入规格：
每个输入文件包含一个测试用例。每个测试用例包含一个不超过20位的正整数。
输出规范：
对于每个测试用例，如果加倍数字由原来的数字组成，那么打印“Yes”否则打印“No”然后再下一行打印出加倍的数字。

代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int i,j,n,m,k,t,flag=0;
    char str[21],result[30];
    int arr[10],brr[10];
    memset(arr,0,sizeof(arr));
    memset(brr,0,sizeof(arr));
    scanf("%s",str);
    for(i=0;str[i]!='\0';i++)
    {
        arr[str[i]-'0']++;
    }
    k=0;
    for(i=strlen(str)-1,j=0;i>=0;i--)
    {
       n=str[i]-'0';
       m=n*2;
       m+=k;
       k=m/10;
       result[j++]=m%10+'0';
    }
    if(k!=0)
    {
        result[j++]=k+'0';
    }
    result[j]='\0';
    for(i=0;result[i]!='\0';i++)
    {
        brr[result[i]-'0']++;
    }
    for(i=0;i<10;i++)
    {
        if(arr[i]!=brr[i])
            break;
    }
    if(i==10)
    {
        printf("Yes\n");
    }
    else
    {
        printf("No\n");
    }
    for(i=strlen(result)-1;i>=0;i--)
        printf("%c",result[i]);
    return 0;
}