UVA 465 Overflow

最新推荐文章于 2019-02-16 22:23:44 发布

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最新推荐文章于 2019-02-16 22:23:44 发布

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分类专栏：模拟

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本文介绍了一个程序设计问题，即如何判断两个非负整数通过加法或乘法运算后的结果是否会超出标准整型变量的表示范围。文章提供了一段示例代码，该代码能够读取输入的表达式，并判断其是否会产生溢出。

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Overflow

Write a program that reads an expression consisting of two non-negative integer and an operator. Determine if either integer or the result of the expression is too large to be represented as a ``normal'' signed integer (type integer if you are working Pascal, type intif you are working 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 containing as 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

题意：

给一个表达式，判断这两个数与表达式的结果是否有大于 int 最大值的，

刚开始理解错了题意，WA好多次！double 保存结果，不超。

代码：

#include <cstdio>
#include <cstring>
#include<climits>       // INT_MAX 需要

int main()
{
    char s1[1000], s2[1000], ch;
    int la, lb;
    while( ~scanf("%s %c %s", &s1, &ch, &s2)  ) {
        double sum1 = 0, sum2 = 0;
        la = strlen(s1);
        lb = strlen(s2);
        for(int i = 0; i < la; i++) sum1 = 10 * sum1 + s1[i] - '0';
        for(int i = 0; i < lb; i++) sum2 = 10 * sum2 + s2[i] - '0';
        printf("%s %c %s\n", s1, ch, s2);
        if(sum1 > INT_MAX) printf("first number too big\n");
        if(sum2 > INT_MAX) printf("second number too big\n");
        if(ch == '+') sum1 += sum2 ;
        if(ch == '*') sum1 *= sum2 ;
        if(sum1 > INT_MAX) printf("result too big\n");
    }
    return 0;
}