lightoj 1043 - Triangle Partitioning 【水题】

最新推荐文章于 2017-06-13 12:32:00 发布

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最新推荐文章于 2017-06-13 12:32:00 发布

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本文介绍了一种基于三角形面积比求解特定边长的方法。针对已知三角形两边及夹角的情况，通过输入三角形的AB、AC、BC边长及三角形ADE与四边形BDEC的面积比，利用面积比等于边长比的平方这一关键信息，计算出AD的具体长度。

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1043 - Triangle Partitioning

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

See the picture below.

You are given AB, AC and BC. DE is parallel to BC. You are also given the area ratio between ADE and BDEC. You have to find the value of AD.

Input

Input starts with an integer T (≤ 25), denoting the number of test cases.

Each case begins with four real numbers denoting AB, AC, BC and the ratio of ADE and BDEC (ADE / BDEC). You can safely assume that the given triangle is a valid triangle with positive area.

Output

For each case of input you have to print the case number and AD. Errors less than 10^-6 will be ignored.

Sample Input

Output for Sample Input

100 100 100 2

10 12 14 1

7 8 9 10

8.134 9.098 7.123 5.10

Case 1: 81.6496580

Case 2: 7.07106781

Case 3: 6.6742381247

Case 4: 7.437454786

题意：给定AB，AC，BC以及三角形ADE和四边形BCDE的面积比，问你AD的长。

思路：面积比 = 边长之比平方。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define INF 0x3f3f3f
#define eps 1e-4
#define MAXN (100+10)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int main()
{
    int t, kcase = 1; Ri(t);
    W(t)
    {
        double a, b, c, d;
        Rf(a); Rf(b); Rf(c); Rf(d);
        double ans = a * sqrt(d / (d+1));
        printf("Case %d: %.8lf\n", kcase++, ans);
    }
    return 0;
}