#10440: 最後一個側資點過不了 求解QQ


a24230928 (U2FsdGVkX187J7YvyTNp5hNzHxI1zu...)


#include<cstdio>

#include<cmath>

int main()

{

int n,a,b,c,root1,root2,j;

while((scanf("%d%d%d",&a,&b,&c))!=EOF)

{

if((b*b-4*a*c)<0)

printf("No real root\n");

else if((b*b-4*a*c)==0)

{

root1=(-b-sqrt(b*b-4*a*c))/2*a;

printf("Two same roots x=%d\n",root1);

}

else if((b*b-4*a*c)>0)

{

root1=(-b+sqrt(b*b-4*a*c))/2*a;

root2=(-b-sqrt(b*b-4*a*c))/2*a;

if(root1>root2)

printf("Two different roots x1=%d , x2=%d\n",root1,root2);

else

printf("Two different roots x1=%d , x2=%d\n",root2,root1);

}

}

return 0;

}

#10470: Re:最後一個側資點過不了 求解QQ


lemontammy (跩比豬)


#include

#include

int main()

{

int n,a,b,c,root1,root2,j;

while((scanf("%d%d%d",&a,&b,&c))!=EOF)

{

if((b*b-4*a*c)<0)

printf("No real root\n");

else if((b*b-4*a*c)==0)

{

root1=(-b-sqrt(b*b-4*a*c))/2*a;

printf("Two same roots x=%d\n",root1);

}

else if((b*b-4*a*c)>0)

{

root1=(-b+sqrt(b*b-4*a*c))/2*a;

root2=(-b-sqrt(b*b-4*a*c))/2*a;

if(root1>root2)

printf("Two different roots x1=%d , x2=%d\n",root1,root2);

else

printf("Two different roots x1=%d , x2=%d\n",root2,root1);

}

}

return 0;

}


2*a要括弧 變成 (2*a)

否則算是會變為 先/2再*a

#10552: Re:最後一個側資點過不了 求解QQ


sandy010337 (sandy)


#include

#include

int main()

{

int n,a,b,c,root1,root2,j;

while((scanf("%d%d%d",&a,&b,&c))!=EOF)

{

if((b*b-4*a*c)<0)

printf("No real root\n");

else if((b*b-4*a*c)==0)

{

root1=(-b-sqrt(b*b-4*a*c))/2*a;

printf("Two same roots x=%d\n",root1);

}

else if((b*b-4*a*c)>0)

{

root1=(-b+sqrt(b*b-4*a*c))/2*a;

root2=(-b-sqrt(b*b-4*a*c))/2*a;

if(root1>root2)

printf("Two different roots x1=%d , x2=%d\n",root1,root2);

else

printf("Two different roots x1=%d , x2=%d\n",root2,root1);

}

}

return 0;

}


2*a要括弧 變成 (2*a)

否則算是會變為 先/2再*a

  真是非常感謝!!卡在這個點有點困擾^_^