Question

题目18：用C语言设计一个简单的计算器，要求能够对输入的数 1.进行+，-，*，/，运算； 2.可以带括号( )；

要正确的答案，错的别来

Answer 1

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int deal(int op1,int op2,char op) //运算
{
switch(op)
{
case '+': op1 += op2; break;
case '-': op1 -= op2; break;
case '*': op1 *= op2; break;
case '/': op1 /= op2; break;
case '^':
{
int temp = op2;
for(int i = 1; i < op2 - 1; i++)
op1 *= temp;
break;
}
}
return op1;
}
char *replace(char *source, char *sub, char *rep) //字符串替换
{
char *result;
char *pc1, *pc2, *pc3;
int isource, isub, irep;
isub = strlen(sub);
irep = strlen(rep);
isource = strlen(source);
if(NULL == *sub)
return strdup(source);
result = (char *)malloc(( (irep > isub) ? (float)strlen(source) / isub* irep+ 1:isource ) * sizeof(char));
pc1 = result;
while(*source != NULL)
{
pc2 = source;
pc3 = sub;
while(*pc2 == *pc3 && *pc3 != NULL && *pc2 != NULL)
pc2++, pc3++;
if(NULL == *pc3)
{
pc3 = rep;

while(*pc3 != NULL)
*pc1++ = *pc3++;
pc2--;
source = pc2;
}
else
*pc1++ = *source;
source++;
}
*pc1 = NULL;
return result;
}

int main()
{
char s[1000],op1[100],op2[100],*p,*temp,*back;
scanf("%s",&s);
temp = s;
back = s;
int pos = -1;
int fin,oper1,oper2,tmp, size;
char op;
while(1)
{
size = (int)strlen(temp);
for(int i = 0; i < size; i++,temp++)
{
if( *temp == '(')
{
p = (temp + 1);
pos =(i + 1); //找出最后一个(的位置
}
}
int k = 0;
if(pos != -1)
{
temp = back;
tmp = pos;
while(*p != ')')
{
p++;
tmp++;
if( *p == '*' || *p == '/')
{
tmp--;
for(;(temp[tmp] >= '0' && temp[tmp] <= '9') || temp[tmp] == '.';)
{
tmp--;
break;
}
}
else
continue;
}
if(tmp != pos && *p != ')')
{
pos = ++tmp;
}
int start = pos;
while( temp[pos] >= '0' && temp[pos] <= '9')
{
op1[k] = temp[pos];
pos++;
k++;
}
op1[k] = '\0';
while((temp[pos] >= '0' && temp[pos] <= '9') || temp[pos] == '.')
pos++;
op = temp[pos];
oper1 = atoi(op1);
pos++;
k = 0;
while( temp[pos] >= '0' && temp[pos] <= '9')
{
op2[k] = temp[pos];
pos++;
k++;
}
op2[k] = '\0';
while((temp[pos] >= '0' && temp[pos] <= '9') || temp[pos] == '.')
pos++;
oper2 = atoi(op2);
int end;
if( temp[pos] != ')')
end = --pos;
else
{
if(temp[start-1] == '(')
{
end = pos;
start--;
}
}
char rep[100];
k = 0;
for(; start <= end; start++,k++)
{
rep[k] = temp[start];
}
rep[k] = '\0';
fin = deal(oper1,oper2,op);
char final[100];
itoa(fin,final,10);
temp = replace(temp,rep,final);
back = temp;
pos = -1;
p = temp;
k = 0;
*rep = NULL;
*final = NULL;
printf("%s\n",replace(temp,rep,final));
}
else
{
temp = back;
p = temp;
tmp = pos;
while(*p != '\0')
{
p++;
tmp++;
if( *p == '*' || *p == '/')
{
tmp--;
for(;(temp[tmp] >= '0' && temp[tmp] <= '9') || temp[tmp] == '.';)
{
tmp--;
break;
}
}
else
continue;
break;
}
if(tmp != pos && tmp != (int)strlen(temp)-1)
{
pos = ++tmp;
}
else
pos = 0;
int start = pos;
while( temp[pos] >= '0' && temp[pos] <= '9')
{
op1[k] = temp[pos];
pos++;
k++;
}
while((temp[pos] >= '0' && temp[pos] <= '9') || temp[pos] == '.')
pos++;
op1[k] = '\0';
op = temp[pos];
oper1 = atoi(op1);
pos++;
k = 0;
while( temp[pos] >= '0' && temp[pos] <= '9')
{
op2[k] = temp[pos];
pos++;
k++;
}
op2[k] = '\0';
while((temp[pos] >= '0' && temp[pos] <= '9') || temp[pos] == '.')
{
pos++;
if(pos >= (int)strlen(temp))
break;
}
oper2 = atoi(op2);
int end = --pos;
char rep[100] = {'\0'};
k = 0;
for(; start <= end; start++)
{
rep[k] = temp[start];
k++;
}
fin = deal(oper1,oper2,op);
char final[100];
itoa(fin,final,10);
temp = replace(temp,rep,final);
back = temp;
for(k = 0; k < (int)strlen(temp); k++)
{
if(temp[k] < '0' || temp[k] > '9' && temp[k] != '.')
break;
}
if(k == (int)strlen(temp))
break;
pos = -1;
p = temp;
k = 0;
printf("%s\n", temp);
}
}
printf("%d\n",fin);
return 0;
}

Answer 2

#include <stdio.h>
int main(）
{
int a,b;
char op;
scanf("%d%c%d",&a,&op,&b);
switch(op)
{
case'+':
printf("%d+%d=%d\n",a,b,a+b);
break;
case'-':
printf("%d-%d=%d\n",a,b,a-b);
break;
case'*':
printf("%d*%d=%d\n",a,b,a*b);
break;
case'/':
if(0==b)
{
printf("Division by zero!\n");
}
else
{
printf("%d/%d=%d\n",a,b,a/b);
}
break;
}
return 0;
}