Overflow 

Write a program that reads an expression consisting of twonon-negative integer and an operator. Determine if either integer orthe result of the expression is too large to be represented as a``normal'' signed integer (typeinteger if you are workingPascal, type int if you are working in C).

Input

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

Output

For each line of input, print the input followed by 0-3 linescontaining as many of these three messages as are appropriate: ``firstnumber 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:输入num1 + 或 * num2

若num1大于int可表示的最大值, 那么输出"first number too big"

同理num2的话, 输出"second number too big"

最后还要判定结果是否溢出, 若溢出, 输出"result too big"

都没溢出, 那么就没输出~

AC代码:

#include<stdio.h>
#include<stdlib.h> #define MAX 2147483647 int main() {
char num1[600], num2[600];
char ch;
double n1, n2;
while(scanf("%s %c %s", num1, &ch, num2) != EOF) {
printf("%s %c %s\n", num1, ch, num2);
n1 = atof(num1);
n2 = atof(num2);
if(n1 > MAX)
printf("first number too big\n");
if(n2 > MAX)
printf("second number too big\n");
if(ch == '+' && n1+n2 > MAX)
printf("result too big\n");
if(ch == '*' && n1*n2 > MAX)
printf("result too big\n");
}
return 0;
}

UVA 465 (13.08.02)的更多相关文章

  1. UVA 10494 (13.08.02)

    点此连接到UVA10494 思路: 采取一种, 边取余边取整的方法, 让这题变的简单许多~ AC代码: #include<stdio.h> #include<string.h> ...

  2. UVA 424 (13.08.02)

     Integer Inquiry  One of the first users of BIT's new supercomputer was Chip Diller. Heextended his ...

  3. UVA 10106 (13.08.02)

     Product  The Problem The problem is to multiply two integers X, Y. (0<=X,Y<10250) The Input T ...

  4. UVA 10194 (13.08.05)

    :W Problem A: Football (aka Soccer)  The Problem Football the most popular sport in the world (ameri ...

  5. UVA 253 (13.08.06)

     Cube painting  We have a machine for painting cubes. It is supplied withthree different colors: blu ...

  6. UVA 573 (13.08.06)

     The Snail  A snail is at the bottom of a 6-foot well and wants to climb to the top.The snail can cl ...

  7. UVA 10499 (13.08.06)

    Problem H The Land of Justice Input: standard input Output: standard output Time Limit: 4 seconds In ...

  8. UVA 10025 (13.08.06)

     The ? 1 ? 2 ? ... ? n = k problem  Theproblem Given the following formula, one can set operators '+ ...

  9. UVA 536 (13.08.17)

     Tree Recovery  Little Valentine liked playing with binary trees very much. Her favoritegame was con ...

随机推荐

  1. Winfroms---看看吧客官~

    假 如 你 的 人 生 有 理 想,那 么 就 一 定 要 去 追,不 管 你 现 在 的 理 想 在 别 人 看 来是 多 么 的 可 笑 , 你 也 不 用 在 乎 , 人 生 蹉 跎 几 十 年 ...

  2. jquery 过滤器

    1.基本选择器 基本选择器是JQuery中最常用的选择器,也是最简单的选择器,它通过元素id.class 和标签名来查找DOM元素.这个非常重要,下面的内容都是以此为基础,逐级提高的. 1).“$(“ ...

  3. [转载] java中byte数组与int,long,short间的转换

    文章转载自http://blog.csdn.net/leetcworks/article/details/7390731 package com.util; /** * * <ul> * ...

  4. Win32中GDI+应用(一)

    GDI+, Microsoft Graphics Device Interface Plus, 是微软在继GDI(Microsoft Graphics Device Interface)后推出的图形编 ...

  5. hdu 1573 A/B (扩展欧几里得)

    Problem Description 要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973)= 1). Input 数据的第一行 ...

  6. 编写类String的构造函数、拷贝构造函数、析构函数和赋值函数

    一.题目: class String { public: String(const char *str = NULL); // 普通构造函数 String(const String &othe ...

  7. PHP 编译问题PEAR package PHP_Archive not installed的解决

    php 的编译时需要依赖pear package ,目前的问题错误"PEAR package PHP_Archive not installed",已经明显报出这个问题. 因此编译 ...

  8. JS中关于clientWidth offsetWidth srollWidth等的含义

    网页可见区域宽: document.body.clientWidth;网页可见区域高: document.body.clientHeight;网页可见区域宽: document.body.offset ...

  9. javaScript常用方法整合(项目中用到过的)

    防止输入空格.缩进等字符: function trim(str){ return str.replace(/^\s+|\s+$/g,""); } JS去掉style样式标签 fun ...

  10. Spring面试笔记

    1. Spring工作机制及为什么要用?Spring 是一个开源框架,是为了解决企业应用程序开发复杂性而创建的.Spring既是一个AOP框架,也是一IOC容器.SpringFramework的组成: ...