UVa 465 Overflow (比较大小难道不应该用double吗)

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本文提供了一道关于整数溢出判断的程序设计题目解答。任务要求读取两个非负整数及一个操作符，判断输入的整数或运算结果是否超出标准整型变量的表示范围。通过解析输入数据并进行相应的加法或乘法运算来实现这一功能。

465 - Overflow

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=97&page=show_problem&problem=406

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 int if 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

如标题所述。

完整代码：

/*0.019s*/

#include<cstdio>
const double maxint = -1u >> 1;

char str[600];

int main()
{
	double a, b;
	char op;
	while (gets(str))
	{
		puts(str);
		sscanf(str, "%lf %c %lf", &a, &op, &b);
		if (a > maxint) puts("first number too big");
		if (b > maxint) puts("second number too big");
		if (op == '+' && a + b > maxint || op == '*' && a * b > maxint) puts("result too big");
	}
	return 0;
}