ZOJ 1241 Geometry Made Simple
2015-07-21 23:45
375 查看
ZOJ Problem Set - 1241
Geometry Made Simple
Time Limit: 2 Seconds
Memory Limit: 65536 KB
Mathematics can be so easy when you have a computer. Consider the following example. You probably know that in a right-angled triangle, the length of the three sides a, b, c (where c is the longest side, called the hypotenuse) satisfy the relation a*a+b*b=c*c.
This is called Pythagora's Law.
Here we consider the problem of computing the length of the third side, if two are given.
Input
The input contains the descriptions of several triangles. Each description consists of a line containing three integers a, b and c, giving the lengths of the respective sides of a right-angled triangle. Exactly one of the three numbers is equal
to -1 (the 'unknown' side), the others are positive (the 'given' sides).
A description having a=b=c=0 terminates the input.
Output
For each triangle description in the input, first output the number of the triangle, as shown in the sample output. Then print "Impossible." if there is no right-angled triangle, that has the 'given' side lengths. Otherwise output the length
of the 'unknown' side in the format "s = l", where s is the name of the unknown side (a, b or c), and l is its length. l must be printed exact to three digits to the right of the decimal point.
Print a blank line after each test case.
Sample Input
3 4 -1
-1 2 7
5 -1 3
0 0 0
Sample Output
Triangle #1
c = 5.000
Triangle #2
a = 6.708
Triangle #3
Impossible.
注:此题为:ZOJ Problem Set - 1241 Geometry Made Simple
说明:注意三角形的条件 以及输出格式
已AC源代码:
#include<stdio.h>
#include<math.h>
int main()
{
int A,B,C,k=1;
while(scanf("%d%d%d",&A,&B,&C)&&(A!=0||B!=0||C!=0))
{
double a,b,c;
a=A;b=B;c=C;
printf("Triangle #%d\n",k++);
if(a==-1)
{
if(c<b)
printf("Impossible.\n");
else
{
a=sqrt(c*c-b*b);
printf("a = %.3lf\n",a);
}
}
if(b==-1)
{
if(c<a)
printf("Impossible.\n");
else
{
b=sqrt(c*c-a*a);
printf("b = %.3lf\n",b);
}
}
if(c==-1)
{
c=sqrt(a*a+b*b);
printf("c = %.3lf\n",c);
}
printf("\n");
}
return 0;
}
Geometry Made Simple
Time Limit: 2 Seconds
Memory Limit: 65536 KB
Mathematics can be so easy when you have a computer. Consider the following example. You probably know that in a right-angled triangle, the length of the three sides a, b, c (where c is the longest side, called the hypotenuse) satisfy the relation a*a+b*b=c*c.
This is called Pythagora's Law.
Here we consider the problem of computing the length of the third side, if two are given.
Input
The input contains the descriptions of several triangles. Each description consists of a line containing three integers a, b and c, giving the lengths of the respective sides of a right-angled triangle. Exactly one of the three numbers is equal
to -1 (the 'unknown' side), the others are positive (the 'given' sides).
A description having a=b=c=0 terminates the input.
Output
For each triangle description in the input, first output the number of the triangle, as shown in the sample output. Then print "Impossible." if there is no right-angled triangle, that has the 'given' side lengths. Otherwise output the length
of the 'unknown' side in the format "s = l", where s is the name of the unknown side (a, b or c), and l is its length. l must be printed exact to three digits to the right of the decimal point.
Print a blank line after each test case.
Sample Input
3 4 -1
-1 2 7
5 -1 3
0 0 0
Sample Output
Triangle #1
c = 5.000
Triangle #2
a = 6.708
Triangle #3
Impossible.
注:此题为:ZOJ Problem Set - 1241 Geometry Made Simple
说明:注意三角形的条件 以及输出格式
已AC源代码:
#include<stdio.h>
#include<math.h>
int main()
{
int A,B,C,k=1;
while(scanf("%d%d%d",&A,&B,&C)&&(A!=0||B!=0||C!=0))
{
double a,b,c;
a=A;b=B;c=C;
printf("Triangle #%d\n",k++);
if(a==-1)
{
if(c<b)
printf("Impossible.\n");
else
{
a=sqrt(c*c-b*b);
printf("a = %.3lf\n",a);
}
}
if(b==-1)
{
if(c<a)
printf("Impossible.\n");
else
{
b=sqrt(c*c-a*a);
printf("b = %.3lf\n",b);
}
}
if(c==-1)
{
c=sqrt(a*a+b*b);
printf("c = %.3lf\n",c);
}
printf("\n");
}
return 0;
}
相关文章推荐
- LeetCode Valid Sudoku
- convertView,这个是Android在为我们而做的缓存机制。
- 按照规定删除字符串
- LeetCode Valid Sudoku
- .Net学习笔记----2015-07-21(C#基础复习07,关键字、访问修饰符)
- GIT入门篇-基本概念与操作
- 关于Map集合的遍历总结
- 一步一步学习网页开发之一
- 设计模式(6) - 创建型模式总结
- Coins
- 红黑树
- PBKDF2加密
- 任务,信号量、邮箱、队列及其区别
- 第63讲Scala中隐式参数与隐式转换的联合使用解析
- 【数据库开发】认识事务
- java 去除url获取值里的BOM
- Snappy压缩
- How to Cite a Patent
- 【求出所有最短路+最小割】【多校第一场】【G题】
- 【求出所有最短路+最小割】【多校第一场】【G题】