1817: Triangle
| Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
|---|---|---|---|---|---|
| 5s | 8192K | 2409 | 685 | Standard |
2nd JOJ Cup Online VContest Problem
Given three integers a, b and c(|a|, |b|, |c|<10000), determine if they can compose a valid triangle. If so, further determine which kind of the triangle it is.
Input Specification
The input consists of several lines, each of which contains three integers a, b and c.
Output Specification
For each group of a b and c, first print "Case K:", a space, and then one of the following four lines:
is not a valid triangle.
is a sharp triangle.
is a right triangle.
is a blunt triangle.
corresponding to your determinant, where K is the number of the test cases starting from 1.
Sample Input
3 4 5
3 3 3
1 2 3
3 4 6
Sample Output
Case 1: is a right triangle.
Case 2: is a sharp triangle.
Case 3: is not a valid triangle.
Case 4: is a blunt triangle.
#include <iostream>
using namespace std;
int a,b,c,tmp;
int main()
{
int count = 0;
//while(scanf("%d%d%d",&a,&b,&c),a,b,c)
while(cin>>a>>b>>c)
{
if(a < b)
{
tmp = a;
a = b;
b = tmp;
}
if(a < c)
{
tmp = a;
a = c;
c = tmp;
}
count ++ ;
if(a >= (b + c))
{
printf("Case %d is not a valid triangle.\n",count);
continue;
}
if(a*a == b*b + c*c)
printf("Case %d is a right triangle.\n",count);
else if(a*a > b*b + c*c)
printf("Case %d is a blunt triangle.\n",count);
else if(a*a < b*b + c*c)
printf("Case %d is a sharp triangle.\n",count);
}
return 0;
}
总是WA一点问题没有啊。。。
+01的正确代码
#include <iostream>
using namespace std;
int main(void){
int a,b,c,r,count=0;
while(cin>>a>>b>>c){
if(a<b){
r = a;
a = b;
b = r;
}
if(a<c){
r = a;
a = c;
c = r;
}
if(b<c){//此处没有必要
r = b;
b = c;
c = r;
}
if(a>=b+c) //偶然的避开了 1 2 3时候 a*a > b* b + c*c
cout<<"Case "<<++count<<": is not a valid triangle."<<endl;
else if(a*a==b*b+c*c)
cout<<"Case "<<++count<<": is a right triangle."<<endl;
else if(a*a>b*b+c*c)
cout<<"Case "<<++count<<": is a blunt triangle."<<endl;
else if(a*a<b*b+c*c)
cout<<"Case "<<++count<<": is a sharp triangle."<<endl;
}
return 0;
}
搞不明白
本文分享了两个C++程序代码实现,用于判断三个整数是否能构成有效的三角形,并进一步确定其为锐角、直角还是钝角三角形。通过对输入数据进行排序和条件判断,实现了对不同类型的三角形的有效识别。
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