杭电2714&&poj2190 ISBN

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ISBN

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1306    Accepted Submission(s): 389


Problem Description
Farmer John's cows enjoy reading books, and FJ has discovered that his cows produce more milk when they read books of a somewhat intellectual nature. He decides to update the barn library to replace all of the cheap romance novels with textbooks on algorithms and mathematics. Unfortunately, a shipment of these new books has fallen in the mud and their ISBN numbers are now hard to read.

An ISBN (International Standard Book Number) is a ten digit code that uniquely identifies a book. The first nine digits represent the book and the last digit is used to make sure the ISBN is correct. To verify that an ISBN number is correct, you calculate a sum that is 10 times the first digit plus 9 times the second digit plus 8 times the third digit ... all the way until you add 1 times the last digit. If the final number leaves no remainder when divided by 11, the code is a valid ISBN.

For example 0201103311 is a valid ISBN, since
10*0 + 9*2 + 8*0 + 7*1 + 6*1 + 5*0 + 4*3 + 3*3 + 2*1 + 1*1 = 55.

Each of the first nine digits can take a value between 0 and 9. Sometimes it is necessary to make the last digit equal to ten; this is done by writing the last digit as X. For example, 156881111X is a valid ISBN number.

Your task is to fill in the missing digit from a given ISBN number where the missing digit is represented as '?'.
 

Input
* Line 1: A single line with a ten digit ISBN number that contains '?' in a single position
 

Output
* Line 1: The missing digit (0..9 or X). Output -1 if there is no acceptable digit for the position marked '?' that gives a valid ISBN.
 

Sample Input
15688?111X
 

Sample Output
1
 

Source
USACO 2003 Fall Orange
 

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我承认做的时候突然傻逼了一下,然后厚着脸皮去问大神,虽然大神并没有给我解答。
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char s[110];int i,j,k,l,m,n;int main(){while(scanf("%s",s)!=EOF){int sum=0;l=strlen(s);for(i=0;i<l;i++){if(s[i]>='0'&&s[i]<='9')sum+=(s[i]-'0')*(10-i);if(s[i]=='X')sum+=10*(10-i);if(s[i]=='?')k=i;}if(k==9){for(i=0;i<11;i++)if((sum+i*(10-k))%11==0){if(i==10)printf("X\n");elseprintf("%d\n",i);break;}if(i==11)printf("-1\n");}else if(k<9){for(i=0;i<10;i++)if((sum+i*(10-k))%11==0){printf("%d\n",i);break;}if(i==10)printf("-1\n");}}}


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