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('')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 = 2147483647;
const int maxn = 300;
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) == 3)
    {
        printf("%s %c %s
", 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 0;
}

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 = 300;
const char *INT = "2147483647";
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, 0, sizeof(x));
            memset(y, 0, sizeof(y));
            for(i = la - 1; i >= 0; --i)
                x[la - 1 - i] = a[i] - '0';
            for(i = lb - 1; i >= 0; --i)
                y[lb - 1 - i] = b[i] - '0';
            for(i = 0; i < lb; ++i)//高精度加法运算
            {
                x[i] = x[i] + y[i];
                if(x[i] > 10)
                {
                    j = i;
                    while(x[j] > 10)//处理连续进位的问题
                    {
                        x[j++] -= 10;
                        ++x[j];
                    }
                }
            }
        }

        if(op == '*')
        {
            //如果乘数为0的话，那么结果不溢出
            if((la == 1 && a[0] == '0') || (lb == 1 && b[0] == '0'))
                continue;
            else
            {
                if((flaga || flagb) || (la + lb > 11))//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, 0, sizeof(x));
                memset(y, 0, sizeof(y));
                for(i = la - 1; i >= 0; --i)
                    x[la - 1 - i] = a[i] - '0';
                for(i = lb - 1; i >= 0; --i)
                    y[lb - 1 - i] = b[i] - '0';

                for(i = 0; i < lb; ++i)
                {
                    int s = 0, c = 0;
                    for(j = 0; j < maxn; ++j)
                    {
                        s = x[j] * y[i] + c;
                        x[i + j] = s % 10;
                        c = s / 10;
                    }
                }
            }
        }

        for(i = maxn - 1; i >= 0; --i)
            if(x[i] != 0)
                break;
        memset(result, 0, sizeof(result));
        j = 0;
        for(; i >= 0; --i)
            result[j++] = x[i] + '0';//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 0;
}
//判断一个字符串是否会溢出
bool judge(char c[], int l)
{
    if(l > 10)
        return true;
    if(l < 10)
        return false;
    if(strcmp(c, INT) > 0)
        return true;
    return false;
}
//忽略字符串的前导0
void fun(char c[])
{
    if(c[0] != '0')
        return;
    int i = 0, j = 0;
    while(c[i] == '0')
        ++i;
    for(; c[i] != ''; ++i)
        c[j++] = c[i];
    c[j] = '';
}