【c++】PAT (Advanced Level)1023. Have Fun with Numbers (20)

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

终于有一道一次AC

#include <iostream>
#include <string>
using namespace std;

string change2s(int a){
	int m;
	string ss,sa="0";
	while(a>0){
		m=a%10;
		sa[0]=m+'0';
		ss.append(sa);
		a=a/10;
	}
	return ss;
}

int main(){
	string a;
	while(cin>>a){
		string b="0",ss;
		int temp=0,digit,n;
		n=a.size();
		for(int i=1;i<=n;i++){
			int sum=(a[n-i]-'0')*2+temp;
			if(sum>9){
				temp=1;
				digit=sum-10+'0';
			}else{
				digit=sum+'0';
				temp=0;
			}
			b[0]=digit;
			ss.insert(0,b);
		}
		if(temp==1){
			ss.insert(0,"1");
			cout<<"No"<<endl<<ss;
		}else {
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					if(ss[i]==a[j]){
						a[j]='a';
					}
				}
			}

			bool flag=true;

			for(int i=0;i<n;i++){
				if(a[i]!='a'){
					flag=false;
				}
			}

			if(flag==false){
				cout<<"No";
			}else{
				cout<<"Yes";
			}
			cout<<endl<<ss;

		}
	}
	return 0;
}