1023 Have Fun with Numbers （20 分）(大数相乘)

最新推荐文章于 2022-03-03 20:54:29 发布

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最新推荐文章于 2022-03-03 20:54:29 发布

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本文介绍了一个有趣的问题：如何判断一个数的两倍是否仅由原数的数字重新排列组成。通过解析示例输入和输出，文章详细阐述了算法设计思路，包括如何处理多位数和进位情况，以及如何验证结果的有效性。

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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

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
using namespace std;
int book[10];
int main(){
    string num;
    cin>>num;
    int flag=0;
    for(int i=num.length()-1;i>=0;i--){
        int temp=num[i]-'0';
        book[temp]++;
        temp=temp*2+flag;
        flag=0;
        if(temp>=10){
            temp-=10;
            flag=1;
        }
        num[i]=temp+'0';
        book[temp]--;
    }
    int flag1=0;
    for(int i=0;i<10;i++)
        if(book[i]) flag1=1;
    if(flag||flag1) cout<<"No\n";
    else cout<<"Yes\n";
    if(flag==1) cout<<"1";
    cout<<num;
    return 0;
}