It is well known that you cannot make a triangle with non-zero area whose sides have lengths 1, 2, 3. Can you make a tetrahedron(四面体) with non-zero volume whose edges have lengths 1, 2, 3, 4, 5, 6?
地大邀请赛d
Problem D: Tetrahedron Inequality
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 15 Solved: 3
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Description
Input
The first line of input contains an integer 0 < n <= 10000, the number of lines to follow. Each of the next n lines contains six positive integers separated by spaces, the lengths of the edges of the desired tetrahedron. The length of each edge is no greater than 1000000.
Output
Output n lines, each containing the word YES if it is possible to construct a tetrahedron with non-zero volume with the given edge lengths, or the word NO if it is not possible.
Sample Input
21 2 3 4 5 610 10 10 10 10 18
Sample Output
NONO
队友所写，之后，为这题不知争了多少次，调了不知道多长时间，怎么测都对，提交就是不对，最后原来这题
还是那个精度问题，因为，有可能是1000000,这样肯定，越界了的，所以必须太大的时候，除个１０００才行，警记
#include<stdio.h>
#include<math.h>
int judge(double a,double b,double c) //判定能否组成三角形
{
if(a+b>c&&a+c>b&&b+c>a)
return 1;
else return 0;
}
int cal(double a,double b,double c,double d,double e,double f)
{
if(a>1000)
{
a/=1000.0;
b/=1000.0;
c/=1000.0;
d/=1000.0;
e/=1000.0;
f/=1000.0;
}
double L1,L2,h1,h2,x1,x2,ff;
L1=(a+b+d)/2.0;
L2=(a+c+e)/2.0;
h1=2*sqrt(L1)*sqrt(L1-a)/a*sqrt(L1-b)*sqrt(L1-d);//
h2=2*sqrt(L2)*sqrt(L2-a)/a*sqrt(L2-c)*sqrt(L2-e);
x1=(b*b+e*e-d*d-c*c)/4/a/a*(b*b+e*e-d*d-c*c);//是平方过的,
//x2=(b*b+e*e-d*d-c*c)*(b*b+e*e-d*d-c*c)/4/a/a;
ff=f*f;
if(ff<(h1+h2)*(h1+h2)+x1&&ff>(h1-h2)*(h1-h2)+x1)
return 1;
else return 0;
}
int main()
{
int t,a,b,c,d,e,f,i,j;
double r[6];
scanf("%d",&t);
while(t--)
{
for(i=0;i<6;i++)
scanf("%lf",&r[i]);
for(j=0,a=0;a<5;a++)
{
for(b=0;b<5;b++)
{
if(a!=b)
for(c=0;c<5;c++)
{if(a!=c&&b!=c)
for(d=0;d<5;d++)
{if(a!=d&&d!=c&&b!=d&&judge(r[a],r[b],r[d]))
for(e=0;e<5;e++)
{if(a!=e&&d!=e&&b!=e&&c!=e)
if(judge(r[a],r[c],r[e]))
{
if(cal(r[a],r[b],r[c],r[d],r[e],r[5]))
{
j++;
break;
}
}
if(j)
break;
}
if(j)
break;
}
if(j)
break;
}
if(j)
break;
}
if(j)
break;
}
if(j)
printf("YES\n");
else printf("NO\n");
}
return 0;
}