1023 Have Fun with Numbers

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三块不一样的石头

最新推荐文章于 2025-07-28 17:30:49 发布

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本文介绍了一种通过数组模拟的方法来验证一个特定整数，在翻倍后其位数是否为原数位的一个不重复排列的问题。该方法适用于不超过20位的整数，并提供了完整的C/C++代码实现。

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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 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

给你⼀个⻓度不超过20的整数N，求这个整数两倍后的数位是否为原数位的⼀个不重复排列。是输出Yes,否输出N哦，最后换行输出N乘二后的积

思路

数组模拟即可

C/C++

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string s;
    int nums[22]={0},len=0,result[10]={0};
    cin >> s;
    for(int z=s.size()-1;z>=0;z--){
        int num = s[z]-48;
        result[num]++;
        nums[len] += num * 2;
        if(nums[len]>9){
            nums[len+1] = 1;
            nums[len]%=10;
        }
        result[nums[len]]--;
        len++;
    }
    bool flag = true;
    if(nums[len]!=0) flag = false , len++;
    for(int z : result) if(z!=0) flag = false;
    puts((flag?"Yes":"No"));
    for(int z=len-1;z>=0;z--) cout << nums[z];
    return 0;
}