1023 Have Fun with Numbers

Posted 2021-01-01 kkmjy

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

题意：

给定一个正整数（最多有20位），假设这个数是a1a2a3...ak，把这个数乘以2倍，看得到的结果是否仍然是a1a2a3...ak的一个排列

 

思路：

大整数与int型整数的乘法；简单哈希。

代码：

#include <cstdio>
#include <cstring>

struct bign{
    int digit[30];
    int len;
    bign(){
        memset(digit,0,sizeof(digit));
        len=0;
    }
};

bign change(char str[])
{
    bign c;
    c.len=strlen(str);
    int n=c.len;
    for(int i=0;i<n;i++){
        c.digit[i]=str[n-1-i]-‘0‘;
    }
    return c;
}

bign multiply(bign a,int b)
{
    bign c;
    int carry=0;
    for(int i=0;i<a.len;i++){
        int temp=a.digit[i]*b+carry;
        c.digit[c.len++]=temp%10;
        carry=temp/10;        
    }    
    while(carry>0){
        c.digit[c.len++]=carry%10;
        carry/=10;
    }
    return c;
}

void print(bign a)
{
    for(int i=a.len-1;i>=0;i--)
        printf("%d",a.digit[i]);
}

int main()
{
    char str[30];
    scanf("%s",str);
    int hashtable[10];
    memset(hashtable,0,sizeof(hashtable));
    int len=strlen(str);
    for(int i=0;i<len;i++)
        hashtable[str[i]-‘0‘]++;
    bign a=change(str);
    bign double_a=multiply(a,2);
    for(int i=0;i<double_a.len;i++)
        hashtable[double_a.digit[i]]--;
    bool flag=true;
    for(int i=0;i<10;i++){
        if(hashtable[i]!=0) {
            flag=false;
            break;
        }
    }
    if(flag==true) printf("Yes
");
    else printf("No
");
    
    print(double_a);
    return 0;
}