九度[1103]二次方程计算器 

 1 # include<iostream>
 2 # include<string>
 3 # include<iomanip>
 4 # include<cmath>
 5 using namespace std;
 6 int a=0,b=0,c=0;
 7 void getPar(string s,int flag)
 8 {
 9     if(s=="0") return ;
10     int l=s.size();
11     bool neg=true;
12     int i=0;
13     for(i=0;i<l;i++)
14     {
15         int sum=0;
16         if(s[i]=='-') {neg=false;i++;}
17         else if(s[i]=='+'){ neg=true;i++;}
18         while(s[i]>='0' && s[i]<='9')
19         {
20             sum=sum*10+s[i]-'0';
21             i++;
22         }
23         if(sum==0){ sum=1;}
24         if(!neg) sum=-1*sum;
25         if(i==l)
26         {
27             if(flag) c=c-sum;
28             else c=c+sum;
29         }
30         else if(s[i]=='x')
31         {
32             if(i+1<l && s[i+1]=='^')
33             {
34                 if(flag) a=a-sum;
35                 else a=a+sum;
36                 //a=a+flag?-1*sum:sum;
37                 i=i+2;
38             }
39             else {
40                 //b=b+flag? -1*sum:sum;
41                 if(flag) b=b-sum;
42                 else b=b+sum;
43             }
44         }
45     }
46 }
47 int main(){
48     string s;
49     while(cin>>s){
50         a=b=c=0;
51         int pos=s.find('=');
52         int l=s.size();
53         string s1=s.substr(0,pos);
54         string s2=s.substr(pos+1,l-1);
55         getPar(s1,0);
56         getPar(s2,1);
57         double dot=b*b-4*a*c;
58         //cout<<a<<" "<<b<<" "<<c<<endl;
59         if(dot<0) cout<<"No Solution"<<endl;
60         else
61         {
62             double b1=(-b-sqrt(dot))/(2.0*a);
63             double b2=(-b+sqrt(dot))/(2.0*a);
64             if(b1>b2)
65             {cout<<fixed<<setprecision(2)<<b2<<" "<<b1<<endl;}
66             else cout<<fixed<<setprecision(2)<<b1<<" "<<b2<<endl;
67         }
68     }
69     return 0;
70 }
不知道错那了 标记一下
posted @ 2016-02-04 19:44  dreamer123  阅读(216)  评论(0)  编辑  收藏  举报