大数加法 465 - Overflow

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UVa OJ

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 1. Elementary Problem Solving :: Big Number

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 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 + 39999999999999999999999 + 11

Sample Output

300 + 39999999999999999999999 + 11first number too bigresult too big

把两个数字转换为浮点数，然后和MAXN_int 比较

用到一个atof()函数，可以把字符串直接转换为浮点数。头文件是stdlib.h

一开始字符数组开的太小，wa了一次。。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#define MAXN 2147483647using namespace std;int main (){    double a,b;    char numa[500],numb[500],ch;    while(scanf("%s %c %s",numa,&ch,numb)!=EOF)    {        printf("%s %c %s\n",numa,ch,numb);        a=atof(numa);        b=atof(numb);        if (a>MAXN) cout<<"first number too big"<<endl;        if (b>MAXN) cout<<"second number too big"<<endl;        if (ch=='+' && a+b>MAXN) cout<<"result too big"<<endl;        if (ch=='*' && a*b>MAXN) cout<<"result too big"<<endl;    }}