大数相加+string表示大数 1023 Have Fun with Numbers (20分)

本文链接：https://blog.youkuaiyun.com/BLUEsang/article/details/105842209

探讨一种特殊性质的数字：当这类数字翻倍后，其结果仅由原数字中的数字重新排列组成。通过使用字符串和算法实现这一检查过程，并提供了一个具体的编程示例。

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1023 Have Fun with Numbers (20分)

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

解题
大数相加，用string代替；

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;

int main()
{	
	vector<int> book(10,0);
	string t;
	cin>>t;
	
	for(char i:t){
		book[i-'0']++;
	}
	
	reverse(t.begin(),t.end());
	bool flag=0;
	string r="";
	int temp;
	for(int i=0;i<t.size();i++)
	{
		temp=2*(t[i]-'0');
		if(flag) temp+=1;
		if(temp>=10){
				r+=(temp%10+'0');	
				flag=1;
			}
		else {
			r+=(temp+'0');
			flag=0;
		}
		book[temp%10]--;	
	}
	
	if(flag) {
	r+='1';
	book[1]--;
	}

	reverse(r.begin(),r.end());
	bool q=1;
	for(int i=0;i<10;i++)
		if(book[i]!=0)
			{
				cout<<"No"<<endl;
				q=0;	
      			break;
			}
	if(q)
  cout<<"Yes"<<endl;
	cout<<r;
}