1023. Have Fun with Numbers (20)

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题目：

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

注意：

1、一定要非常小心进位，特别是最高位也需要进位的时候！！！

2、由于这里的输入可能是10^20量级，所以我直接用了字符数组来计算，先将每一位数字乘以2，再判断是否需要进位，并统计0~9每个数字出现的次数是否与输入相同。

代码：

#include<iostream>#include<string>#include<cstring>using namespace std;int main(){//x:input number//x2:double numberchar x[21],x2[21],t;//xdigit:appearence time if each digit from 0 to 9//it is more convenient to compare by using function strcmp than int[]char xdigit[11],x2digit[11];memset(xdigit, '0',sizeof (xdigit)); xdigit[10]='\0';strcpy(x2digit,xdigit);int i=0;cin>>x;while(x[i]!='\0' ){++xdigit[x[i]- '0'];//count each digit for xx2[i]=(x[i]- '0')*2+'0' ;//double each digit++i;}x2[i]= '\0';while(--i){if(x2[i]-'0' >9)//{++x2[i-1];         x2[i]-=10;}++x2digit[x2[i]- '0'];//count each digit for x2 except the first digit}if(x2[0]-'0' >9){ //if the first digit of x2 is more than 9,that means the length of x2 increases by 1 digitcout<< "No"<<endl;x2[0]-=10;cout<<1<<x2<<endl;return 0;}else{++x2digit[x2[0]- '0'];//don't forget the first digit!if(strcmp(xdigit,x2digit)==0)cout<< "Yes"<<endl;else cout<<"No" <<endl;cout<<x2;}return 0;}

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