1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 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:

Yes2469135798

#include<iostream>#include<vector>#include<stdio.h>#include<algorithm>#include<math.h>#include<map>using namespace std;int main(){    string str,res="",s="0";    cin>>str;    int len = str.size();    int cflag=0;//进位    for(int i=len-1;i>=0;--i){       int temp = (str[i]-'0')*2+cflag;       cflag = temp/10;       temp = temp%10;       s[0]=temp+'0';       res = s+res;       if(i==0&&cflag!=0)        res = "1"+res;    }    s = res;    sort(s.begin(),s.end(),[](char a,char b){return a<b;});    sort(str.begin(),str.end(),[](char a,char b){return a<b;});    printf(str==s?"Yes\n":"No\n");    printf("%s\n",res.c_str());    return 0;}

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