UVa 465 Overflow——WA

上次那个大数开方的高精度的题，UVa113 Power of Cryptography，直接两个double变量，然后pow(x, 1 / n)就A过去了。

怎么感觉UVa上高精度的题测试数据不给力啊。。。

话说回来，我写了100+行代码，WA了，后来考虑到要忽略前导0，又WA了，实在不知道哪出问题了。

 Overflow 

Write a program that reads an expression consisting of twonon-negative integer and an operator. Determine if either integer or the resultof the expression is too large to be represented as a ``normal'' signed integer(type integer if you areworking Pascal, type int if you areworking in C).

Input

An unspecified number of lines. Each line will contain an integer,one of the two operators + or *, and another integer.

Output

For each line of input, print the input followed by 0-3 lines containingas many of these three messages as are appropriate: ``first number too big'',``second number too big'',``result too big''.

Sample Input

300 + 3

9999999999999999999999 + 11

Sample Output

300 + 3

9999999999999999999999 + 11

first number too big

result too big

看到好多人用atof()，就是把一个字符串转化成double型数据。

下面是百科内容：

表头文件 #include <stdlib.h>

定义函数 double atof(const char *nptr);

函数说明 atof()会扫描参数nptr字符串，跳过前面的空格字符，直到遇上数字或正负符号才开始做转换，而再遇到非数字或字符串结束时('\0')才结束转换，并将结果返回。参数nptr字符串可包含正负号、小数点或E(e)来表示指数部分，如123.456或123e-2。

返回值 返回转换后的浮点型数。

附加说明 atof()与使用strtod(nptr,(char**)NULL)结果相同。

如果用double的话，直接将数据和2^31-1进行比较判断是否溢出就好了。

这是别人的AC代码。

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

 const int INT = ;
 const int maxn = ;
 char a[maxn], b[maxn], c;

 int main(void)
 {
     #ifdef LOCAL
         freopen("465in.txt", "r", stdin);
     #endif

     double x, y, z;
     while(scanf("%s %c %s", a, &c, b) == )
     {
         printf("%s %c %s\n", a, c, b);
         x = atof(a);
         y = atof(b);
         if(c == '+')
             z = x + y;
         if(c == '*')
             z = x * y;
         if(x > INT)
             cout << "first number too big" << endl;
         if(y > INT)
             cout << "second number too big" << endl;
         if(z > INT)
             cout << "result too big" << endl;
     }
     return ;
 }

代码君

可我还是想把自己WA的代码贴出来，毕竟是花了大心思用心去写的。

而且这个代码也通过了样例测试和我自己能想到的各种极端的情况。

 //#define LOCAL
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 const int maxn = ;
 const char *INT = "";
 char a[maxn], b[maxn], c[maxn], result[maxn];
 int x[maxn], y[maxn], z[maxn];
 bool judge(char c[], int l);
 void fun(char c[]);

 int main(void)
 {
     #ifdef LOCAL
         freopen("465in.txt", "r", stdin);
     #endif

     char op;
     while(gets(c))
     {
         cout << c << endl;
         sscanf(c, "%s %c %s", a, &op, b);
         fun(a);
         fun(b);
         int la = strlen(a);
         int lb = strlen(b);
         bool flaga, flagb;
         if(flaga = judge(a, la))
             cout << "first number too big" << endl;
         if(flagb = judge(b, lb))
             cout << "second number too big" << endl;

         int i, j;
         //对结果是否溢出进行处理
         if(op == '+')
         {
             if(flaga || flagb)//加法运算有一个加数溢出那么结果溢出
             {
                 cout << "result too big" << endl;
                 continue;
             }
             memset(x, , sizeof(x));
             memset(y, , sizeof(y));
             for(i = la - ; i >= ; --i)
                 x[la -  - i] = a[i] - '';
             for(i = lb - ; i >= ; --i)
                 y[lb -  - i] = b[i] - '';
             for(i = ; i < lb; ++i)//高精度加法运算
             {
                 x[i] = x[i] + y[i];
                 if(x[i] > )
                 {
                     j = i;
                     while(x[j] > )//处理连续进位的问题
                     {
                         x[j++] -= ;
                         ++x[j];
                     }
                 }
             }
         }

         if(op == '*')
         {
             //如果乘数为0的话，那么结果不溢出
             if((la ==  && a[] == '') || (lb ==  && b[] == ''))
                 continue;
             else
             {
                 if((flaga || flagb) || (la + lb > ))//有一个乘数溢出或者两乘数位数之和超过11位
                 {
                     cout << "result too big" << endl;
                     continue;
                 }
                 memset(x, , sizeof(x));
                 memset(y, , sizeof(y));
                 for(i = la - ; i >= ; --i)
                     x[la -  - i] = a[i] - '';
                 for(i = lb - ; i >= ; --i)
                     y[lb -  - i] = b[i] - '';

                 for(i = ; i < lb; ++i)
                 {
                     int s = , c = ;
                     for(j = ; j < maxn; ++j)
                     {
                         s = x[j] * y[i] + c;
                         x[i + j] = s % ;
                         c = s / ;
                     }
                 }
             }
         }

         for(i = maxn - ; i >= ; --i)
             if(x[i] != )
                 break;
         memset(result, , sizeof(result));
         j = ;
         for(; i >= ; --i)
             result[j++] = x[i] + '';//将结果转化为字符串好进行判断
         if(judge(result, strlen(result)))
             cout << "result too big" << endl;
     }
     return ;
 }
 //判断一个字符串是否会溢出
 bool judge(char c[], int l)
 {
     if(l > )
         return true;
     if(l < )
         return false;
     if(strcmp(c, INT) > )
         return true;
     return false;
 }
 //忽略字符串的前导0
 void fun(char c[])
 {
     if(c[] != '')
         return;
     int i = , j = ;
     while(c[i] == '')
         ++i;
     for(; c[i] != '\0'; ++i)
         c[j++] = c[i];
     c[j] = '\0';
 }

代码君