浙大 pat 1023题解

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

#include"iostream"
#include "algorithm"
#include "string"
#include<stdlib.h>
using namespace std;

string Double(string str)
{
int sum;
string result="";
int carry = 0;
int n = str.length();
for(int i=n-1;i>=0;i--)
{
sum = (str[i]-'0')*2;
sum += carry;
result.insert(result.begin(),(sum%10)+'0');
carry = sum/10;
}
if(carry)
{
result.insert(result.begin(),carry+'0');
}
return result;
}
int main()
{
string str,dstr,tstr;
cin >> str;
dstr = Double(str);
tstr = dstr;
sort(str.begin(),str.end());
sort(tstr.begin(),tstr.end());
if(str == tstr)
cout << "Yes"<<endl;
else
cout <<"No" <<endl;
cout << dstr<<endl;

return 0;
}

转载于:https://www.cnblogs.com/luxiao/p/pat1023.html