Posted on 2009-10-30 00:36
小強摩羯座 閱讀(265)
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算法編程
#include <stdio.h>
int add(int x,int y) {return x+y;}
int sub(int x,int y) {return x-y;}
int mul(int x,int y) {return x*y;}
int div(int x,int y) {return x/y;}
int (*func[])()={add,sub,mul,div};
int num,curch;
char chtbl[]="+-*/()=";
char corch[]="+-*/()=0123456789";
int getach() {
int i;
while(1) {
curch=getchar();
if(curch==EOF) return -1;
for(i=0;corch[i]&&curch!=corch[i];i++);
if(i<strlen(corch)) break;
}
return curch;
}
int getid() {
int i;
if(curch>='0'&&curch<='9') {
for(num=0;curch>='0'&&curch<='9';getach()) num=10*num+curch-'0';
return -1;
}
else {
for(i=0;chtbl[i];i++) if(chtbl[i]==curch) break;
if(i<=5) getach();
return i;
}
}
int cal() {
int x1,x2,x3,op1,op2,i;
i=getid();
if(i==4) x1=cal(); else x1=num;
op1=getid();
if(op1>=5) return x1;
i=getid();
if(i==4) x2=cal(); else x2=num;
op2=getid();
while(op2<=4) {
i=getid();
if(i==4) x3=cal(); else x3=num;
if((op1/2==0)&&(op2/2==1)) x2=(*func[op2])(x2,x3);
else {
x1=(*func[op1])(x1,x2);
x2=x3;
op1=op2;
}
op2=getid();
}
return (*func[op1])(x1,x2);
}
void main(void) {
int value;
printf("Please input an expression:\n");
getach();
while(curch!='=') {
value=cal();
printf("The result is:%d\n",value);
printf("Please input an expression:\n");
getach();
}
}