Game improvement of smart people: Chapter 3 Lesson 1: ISBN (ISBN code)

Here's a description , The title of this question does not match the test data , therefore —— I changed the title ：

[ Problem description ]

Farmer John My cow likes reading , also Farmer John I found that his cow read some books about Natural Science , Produced more milk . He decided to replace the cheap novel with a book about Natural Science . Unfortunately , Some books fell into the mud , Their ISBN I can't see the code clearly .

ISBN code （ International Standard Book Number ） It is a code composed of ten Arabic numerals , Used to uniquely identify a book . If you want to verify ISBN Is the code correct , You have to multiply the first number 10, The second number multiplies 9, The third number multiplies 8, The fourth number multiplies 7…… Until the last number is multiplied 1, Add up these accumulations , If the number can be 11 to be divisible by （ That is to say 11 Multiple ） Words , Then this is a legal ISBN code .

for instance 0201103311 It's a legal ISBN code , because ：

0×10 + 2×9 + 0×8 + 1×7 + 1×6 + 0×5 + 3×4 + 3×3 + 1×2 + 1×1

The first nine numbers are 0 – 9 Between . occasionally , The last number has to be taken 10, Then let's write the last number in capital X（ It's not called numbers at this time , ha-ha ）, such as 156881111X It's also a legal ISBN code . Your task is to give you a number lost ISBN After a yard , Identify the missing number . For missing numbers “ ？” Express .

[ Input format ]

One 10 It's made up of two numbers ISBN code , It includes “ ？” Represents the missing number .

[ Output format ]

Lost ISBN code （0-9 Or in capitals X）. If it says “ ？” There are no numbers that can make it legal ISBN code , The output -1.

[ sample input ]

02011?3311

[ sample output ]

0

well , And then code ：

#include <iostream>
#include <string>
#include <cstdio>
using namespace std;
int i,k,n,t,s=0;
string a;
bool f=1;
int main()
{
	freopen ("isbn.in","r",stdin);
	freopen ("isbn.out","w",stdout);
	cin >>a;
	for (i=0;i<10;i++)
	{
		if (a[i]=='?')
			k=10-i;
		else if (a[i]>='0' && a[i]<='9')
		{
			s+=(a[i]-48)*(10-i);
		} 
		else
			s+=10;
	}
	t=(s/11+1)*11-s;
	if (s!=0)
	{
		if (s%11==0)
			cout <<0;
		else if (k!=1)
		{
			for (i=0;(t+i*11)/k<10;i++)
				if ((t+i*11)%k==0)
				{
					cout <<(t+i*11)/k;
					f=0;
					break;
				}
			if (f)
				cout <<-1;
		}
		else
		{
			if (t<10)
				cout <<t;
			else if (t==10)
				cout <<"X"; 
		}
	}
	else
		cout <<-1; 
	fclose(stdin);
	fclose(stdout);
	return 0;
}

Okay , The code is easy to understand , Don't forget to like it .