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 + 3 9999999999999999999999 + 11
Sample Output
300 + 3 9999999999999999999999 + 11 first number too big result 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<stdio.h> #include<string.h> #include<stdlib.h> 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<stdio.h> #include<string.h> 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<n;i++) { u=u*10+s[i]-'0'; if(u>max) { printf("first number too big\n"); break; } } for(i=0;i<m;i++) { v=v*10+s1[i]-'0'; if(v>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; }
