hdu Integer Inquiry 解题报告

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1047

题目意思：就是求大整数加法。有多个案例，每个案例之间要输出一个空格。

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 using namespace std;

 const int maxn =  + ;
 char t[maxn];
 int s[maxn], ans[maxn];

 int main()
 {
     int l1, l2, n, i, j, l, len, num;
     scanf("%d", &n);
     while (n--)
     {
         num = ;
         memset(ans, , sizeof(ans));
         while (cin >> t && strcmp(t, ""))
         {
             memset(s, , sizeof(s));
             l1 = strlen(t);
             for (i = ; i < l1; i++)
                 s[i] = t[l1-i-] - '';
             num++;
             if (num == )
             {
                 for (i = ; i < l1; i++)
                     ans[i] = s[i];
                 l2 = l1;
             }
             else
             {
                 len = max(l1, l2);  // l1:s[i] l2:ans[i]
                 l = min(l1, l2);
                 int sum, c = ;
                 for (i = ; i < len; i++)
                 {
                     if (i < l)
                     {
                         sum = s[i] + ans[i] + c;
                         ans[i] = sum % ;
                         c = (sum >  ?   : );
                     }
                     else
                     {
                         if (l1 == len)
                             sum = s[i] + c;
                         else
                             sum  = ans[i] + c;
                         ans[i] = sum % ;
                         c = (sum >  ?   : );
                     }
                 }
                 if (c ==  && i == len)
                 {
                     ans[len] = ;
                     l2 = len + ;
                 }
                 else
                     l2 = len;   // 关键之处！！！！之前不记得更新错了好多次！！！
             }
         }
         if (!num)
             printf("");
         else
         {
             for (i = maxn-; i >= ; i--)
             {
                 if (ans[i])
                     break;
             }
             for ( ; i >= ; i--)
                 printf("%d", ans[i]);
         }
         printf("\n");
         if (n)
             cout << endl;
     }
     return ;
 }

　　忽略格式问题，借鉴了别人的写法

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 using namespace std;

 const int maxn = 1e5 + ;
 char t[maxn];
 int num[maxn];

 void add(char s[])
 {
     int z = ;
     int len = strlen(s);
     for (int i = len-; i >= ; i--)
     {
         num[z] += (s[i] - '');
         num[z+] += num[z] / ;
         num[z] %= ;
         z++;
     }
 }

 int main()
 {
     int i, ok = ;
     memset(num, , sizeof(num));
     while (scanf("%s", t) != EOF && strcmp(t, ""))
         add(t);
     for (i = maxn; i >= ; i--)
     {
         if (num[i])
         {
             ok = ;
             break;
         }
     }
     for ( ; i >=  && ok; i--)
          printf("%d", num[i]);
     if (!ok)
         printf("");
     printf("\n");
     return ;
 }