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  1. #pragma GCC optimize("Ofast,fast-math,unroll-loops,no-stack-protector")
  2. #include <stdio.h>
  3. #define N 1<<24
  4. char buf[N],*p1=buf,*p2=buf,buf2[N],*p3=buf2,*p4=buf2,s,str1[]="考試要加油口屋\n\n",str2[]="請使用配方法解下列一元二次方程式\n\n",*ps;
  5. #define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,N,stdin),p1==p2)?EOF:*p1++)
  6. void inline pc(char k){if(p3==p4){puts(buf2);p3=buf2;}*p3++=k;}
  7. int t=0,n,a,b,c,i,j,pi,flag;
  8. int main(){
  9.  p4+=N;
  10.  for(;(s=gc())!='\n';t+=s-'0')t*=10;
  11.  for(;t--;){
  12.   for(n=0;(s=gc())!='\n';n+=s-'0')n*=10;
  13.   for(ps=str2;*ps!='\0';)pc(*ps++);
  14.   for(i=1;i<=n;++i){
  15.    for(flag=0,a=0;(s=gc())!=' ';){if(s!='-'){a*=10;a+=s-'0';}else flag=1;}if(flag)a=-a;
  16.    for(flag=0,b=0;(s=gc())!=' ';){if(s!='-'){b*=10;b+=s-'0';}else flag=1;}if(flag)b=-b;
  17.    for(flag=0,c=0;(s=gc())!='\n'&&s!=EOF;){if(s!='-'){c*=10;c+=s-'0';}else flag=1;}if(flag)c=-c;
  18.    for(pi=i,j=10;j<=pi;)j*=10;
  19.    for(j/=10;j>0;pi%=j,j/=10)pc((pi/j)+'0');
  20.    pc('.');pc(' ');
  21.    if(a<0){pc('-');a=-a;}
  22.    if(a!=1){
  23.     for(j=10;j<=a;)j*=10;
  24.     for(j/=10;j>0;a%=j,j/=10)pc((a/j)+'0');
  25.    }
  26.    pc('x');pc('^');pc('2');
  27.    if(b!=0){
  28.     if(b<0){pc('-');b=-b;}else pc('+');
  29.     if(b!=1){
  30.      for(j=10;j<=b;)j*=10;
  31.      for(j/=10;j>0;b%=j,j/=10)pc((b/j)+'0');
  32.     }
  33.     pc('x');
  34.    }
  35.    if(c!=0){
  36.     if(c<0){pc('-');c=-c;}else pc('+');
  37.     for(j=10;j<=c;)j*=10;
  38.     for(j/=10;j>0;c%=j,j/=10)pc((c/j)+'0');
  39.    }
  40.    pc('=');pc('0');pc('\n');pc('\n');
  41.   }
  42.   for(ps=str1;*ps!='\0';)pc(*ps++);
  43.  }
  44.  if(p3!=buf2)puts(buf2);
  45.  return 0;
  46. }
Success #stdin #stdout 0s 5460KB
comments (?)
stdin
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1
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1 -2 -3
stdout
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請使用配方法解下列一元二次方程式

1. x^2-2x-3=0

考試要加油口屋
https://ideone.com/ZpZ2VX
language:
C (gcc 8.3)
created:
332 days ago
visibility:
public

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