这个题我前几天只是做了表达式变换部分，今天把求值部分加进去了。1
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| #include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stack>
using namespace std;
typedef struct Tree
{
char c;
char ss[25];
double n;
Tree *liftchild,*rightchild;
}tree;
stack <tree*> ops;
stack <tree*> ovs;
tree *Root;
char s[1005];
int p(char ch)
{
switch(ch)
{
case '*':
case '/':return 2;
case '+':
case '-':return 1;
case '(':
case ')':return 0;
case ';':return -1;
default :printf("%c_error!\n",ch);exit(1);
}
}
tree *_build(char *s)
{
tree *root=new tree;
strcpy(root->ss,s);
root->liftchild=NULL;
root->rightchild=NULL;
return root;
}
tree *build(char ch)
{
tree *root=new tree;
root->c=ch;
root->liftchild=NULL;
root->rightchild=NULL;
return root;
}
double calcu(double n1,char op,double n2)
{
switch(op)
{
case '*':return n1*n2;
case '/':return n1/n2;
case '+':return n1+n2;
case '-':return n1-n2;
default :printf("%c_error!",op);return 0;
}
}
void visit(tree * r)
{
if(r!=NULL)
{
visit(r->liftchild);
visit(r->rightchild);
if(r->liftchild==NULL && r->rightchild==NULL)
{ printf("%s",r->ss); r->n=atof(r->ss);}
else
{ printf("%c",r->c); r->n=calcu(r->liftchild->n,r->c,r->rightchild->n);}
}
}
int main()
{
int i,j;
int N;
scanf("%d",&N);
char w[100];
while(N--)
{
scanf("%s",s);
for(i=0;s[i]!='=';)
{
if(s[i]>='0' && s[i]<='9')
{
j=0;
while((s[i]>='0' && s[i]<='9') || s[i]=='.')
{
w[j++]=s[i++];
}
w[j]='\0';
ovs.push(_build(w));
}
else
{
if(ops.empty() || s[i]=='(' || p(ops.top()->c)<p(s[i]) )
ops.push(build(s[i])),i++;
else if(ops.top()->c=='(' && s[i]==')')
ops.pop(),i++;
else
{
ops.top()->rightchild=ovs.top();ovs.pop();
ops.top()->liftchild=ovs.top();ovs.pop();
Root=ops.top();ops.pop();
ovs.push(Root);
}
}
}
while(ovs.size()!=1)
{
ops.top()->rightchild=ovs.top();ovs.pop();
ops.top()->liftchild=ovs.top();ovs.pop();
Root=ops.top();ops.pop();
ovs.push(Root);
}
Root=ovs.top();ovs.pop();
while(!ops.empty())
ops.pop();
visit(Root);
printf("=\n%.2lf\n\n",Root->n);
}
return 0;
}
|
我觉得不是很难，只要表达式求值你会了，这题也是小意思。