HDU 1002 A + B Problem II

纯粹的大数问题

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1002

感觉代码好龊 QAQ  但是写的应该还是比较好理解的...（面壁...

#include <stdio.h>
#include <string.h>
int main()
{
    char s1[1001], s2[1001];
    int s[1005];
    int i, q, w, n, h;
    int z = 0;
    scanf("%d",&n);
    h = n;
    while( n-- )
    {
           z++;
           scanf("%s%s",s1,s2); //字符串输入 
           q = strlen(s1);
           w = strlen(s2);
           if( q > w )
           {
               for( i = q-1; i >= q-w; i--)
                    s[i] = (s1[i] - '0') + (s2[i-q+w] - '0');//数位对齐相加 （将字符转换成数字） 
               for( i = q-w-1; i >= 0; i--)
                    s[i] = s1[i] - '0';
               for( i = q-1; i > 0; i--)//最高位不需要进位处理 可直接输出 
               {
                    if(s[i]/10 != 0)
                    {
                               s[i-1] = s[i-1] + (s[i]/10); //进位处理 
                               s[i] = s[i]%10; 
                    }
               }
               printf("Case %d:\n",z);
               printf("%s + %s = ",s1,s2);
               for( i = 0; i < q; i++)
                    printf("%d",s[i]);
           }
           else if( q == w)
           {
                for( i = w-1; i >= 0; i--)
                    s[i] = (s1[i] - '0') + (s2[i-q+w] - '0');
                for( i = q-1; i > 0; i--)
                {
                    if(s[i]/10 != 0)
                    {
                               s[i-1] = s[i-1] + (s[i]/10);
                               s[i] = s[i]%10;
                    }
                }
                printf("Case %d:\n",z);
                printf("%s + %s = ",s1,s2);
                for( i = 0; i < q; i++)
                    printf("%d",s[i]);
           }
           else
           {
                for( i = w-1; i >= w-q; i--)
                    s[i] = (s1[i-w+q] - '0') + (s2[i] - '0');
                for( i = w-q-1; i >= 0; i--)
                    s[i] = s2[i] - '0';
                for( i = w-1; i > 0; i--)
                {
                    if(s[i]/10 != 0)
                    {
                               s[i-1] = s[i-1] + (s[i]/10);
                               s[i] = s[i]%10;
                    }
                }
                printf("Case %d:\n",z);
                printf("%s + %s = ",s1,s2);
                for( i = 0; i < w; i++)
                    printf("%d",s[i]);
           }
           if(z != h) 
                printf("\n");
           printf("\n");
    }
    return 0;
}