/************************/
#include<stdio.h>
int main()
{
    long long int X,O;  
    int Y,Z,I,F;
    printf("please input the Universal Product Code:\n");
    scanf("%d%d%d%d%d%d%d%d%d%d%d\n",&X);
    O=X;
    for(Y=2;Y<12;Y++)
    {
        if(Y%2==0)
        {
            if(X%10==0&&X>10)
            {
                X=X/10;
            }
            else if(X%10!= 0&&X>10)
            {
                Z=Z+X%10;
                X=X/10;
            }
            else if(X%10!=0&&X<10)
            {
                Z=Z+X;        
            }
            
        }
        else if (Y%2!=0)
        {
            if(X%10==0)
            {
                X=X/10;
            }
            else if(X%10!=0)
            {
                I=I+X%10;
                X=X/10;
            }
        }
                        
    }
    F=(3*Z-I)%10;
    O=O*10+F;
    printf("The 12bits UPC=%d",O);
    return 0;
     
     
     
   
}
/************************/

首先，要贴题目，以下是我从网上拷贝过来的：

“将减法的结果除以10再取余数” ------ 如果相减的结果是负数怎么办？

从网上找到真正的UPC校验码规则：( 10 - (奇数位上的数字和*3 + 偶数位上的数字和)%10 )%10

例如 63938200039 的校验码为 3
奇数位上的数字和 = 6+9+8+0+0+9 = 32
偶数位上的数字和 = 3+3+2+0+3 = 11
3*32 + 11 = 107
7 + 3 等于10的倍数，所以校验码为 3

#include <stdio.h>

int main( void )
{
    unsigned long long upc;
    if( scanf("%llu",&upc)!=1 || upc>99999999999 )
        return 1;

    unsigned c = 0;
    for( unsigned long long tmp=upc; tmp!=0; tmp/=100 )
        c += tmp%10*3 + tmp%100/10;
    c = (10-c%10)%10;

    printf( "%llu\n", upc*10+c );
}

[此贴子已经被作者于2017-11-10 10:36编辑过]

这个解法好厉害，我没想到这么远。但我还想问下就是for循环中    for（unsigned long long tmp=upc....） 中直接可以表明tmp=upc？

除非你教科书是谭浩强写的，且编译器用的都是Turbo C等上古遗物，或VC等明确说不支持C的编译器，否则这是常识呀！

见C标准文档中的 Iteration statements 项
Iteration statements
iteration-statement:
    for( expression; expression; expression ) statement
    for( declaration; expression; expression ) statement
The declaration part of a for statement shall only declare identifiers for objects having storage class auto or register.

The for statement
The statement
    for( clause-1; expression-2; expression-3 ) statement
……. If clause-1 is a declaration, the scope of any identifiers it declares is the remainder of the declaration and the entire loop, including the other two expressions; ……