29: ISBN number, 29isbn number

29: ISBN number, 29isbn number
29: ISBN number

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Total time limit:

1000 ms

 

Memory limit:

65536kB

Description

Each officially published book corresponds to an ISBN number. The ISBN code consists of nine digits, one-digit identification code, and three separators, the format is as follows: "x-xxx-xxxxx-x", where the symbol "-" is a separator (minus sign on the keyboard), and the last digit is an identifier, for example, 0-670-82162-4 is a standard ISBN code. The first digit of the ISBN code represents the publishing language of the book. For example, 0 represents English, and three digits after the first separator "-" represent the publishing house, for example, 670 represents the Viking Publishing House; the fifth digit after the second separator represents the number of the book in the press; the last digit is the identification code.

The identifier calculation method is as follows:

Multiply the first digit by 1 and the second digit by 2 ...... Similarly, if mod 11 is used, the remainder is the identification code. If the remainder is 10, the identification code is an uppercase letter X. For example, Id 4 of ISBN numbers 0-670-82162-4 is obtained as follows: for the 9 Numbers 067082162, multiply them from left to right by 1, 2 ,..., 9, then sum, that is, 0 × 1 + 6 × 2 + „ + 2 × 9 = 158, and then take Result 4 of 158 mod 11 as the identification code.

Your task is to write a program to determine whether the identifier in the input ISBN number is correct. If it is correct, only "Right" is output. If it is incorrect, the ISBN number that you think is correct is output.

Input

Only one line is a character sequence, indicating the ISBN number of a book (ensure that the input meets the ISBN number format requirements ).

Output

If the ID of the ISBN number is correct, "Right" is output. Otherwise, the correct ISBN number (including the separator "-") is output according to the specified format.

Sample Input

Example #1:0-670-82162-4 example #-670-82162-0

Sample output

Example #1: Right example #-670-82162-4

Source

No. 1 in NOIP2008 semi-finals popularity Group

 1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 int t=0,sum=0; 6 char n[15]; 7 scanf("%s",n); 8 int len=strlen(n); 9 for(int i=0;i<len-1;++i)10 {11 if(n[i]>='0'&&n[i]<='9')sum+=(++t)*(n[i]-'0');12 }13 if(sum%11==n[len-1]-'0'||sum%11==10&&n[len-1]=='X')printf("Right");14 else{15 n[len-1]='\0';16 printf("%s",n);17 if(sum%11!=10)printf("%d",sum%11);18 else printf("X");19 }20 return 0;21 }