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最近新学了一个函数的用法，头文件是 stdlib.h 的 atof() 的函数，它可以把字符串的数字转化为double类型。

题目：

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 (typeintegerif you are working Pascal, typeintif 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

解题思路：就是先判断两个数是否超出范围，然后再判断加和乘结果是否超出范围。

1、atof()函数写法：

#include
   
    #include
    
     #include
     
      long long max=2147483647;int main(){
   	char s[5000],s1[5000],t;	double u,v;	while(scanf("%s %c %s",s,&t,s1)!=EOF)	{
   		printf("%s %c %s\n",s,t,s1);		u=atof(s);		v=atof(s1);		if(u>max)			printf("first number too big\n");		if(v>max)			printf("second number too big\n");		if(t=='+'&&u+v>max)			printf("result too big\n");		else if(t=='*'&&u*v>max)			printf("result too big\n");			}	return 0;}
     
    
   

2、不用atof()函数的代码：

#include
   
    #include
    
     long long max=2147483647;int main(){
   	char s[5000],s1[5000],t;	int n,m,i,j,k;	long long u,v;	while(scanf("%s %c %s",s,&t,s1)!=EOF)	{
   		printf("%s %c %s\n",s,t,s1);		n=strlen(s);		m=strlen(s1);		u=v=0;		for(i=0;i
     
      max)			{
   				printf("first number too big\n");				break;			}		} 		for(i=0;i
      
       max)			{
   				printf("second number too big\n");				break;			}		}		if(t=='+'&&u+v>max)			printf("result too big\n");		else if(t=='*'&&u*v>max)			printf("result too big\n");			}	return 0;}
      
     
    
   

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