Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem C: Triangle Trouble

There is trouble at the triangle factory. The triangle assembler has gone down, so all that has been produced for the day is a bunch of triangle sides. To make the best of this situation, it has been decided to create the triangle with the largest possible area from the available sides, and sell it as a limited edition triangle.

You have been hired to write a program that will determine the area of the limited edition triangle.

Input begins with the number of test cases on its own line. Each test case begins with a positive integer N (3 ≤ N ≤ 10,000), followed by N positive real numbers s_i representing the lengths of the available triangle sides (0 < s_i ≤ 100,000). A single test case may be spread out over several consecutive lines of the input.

For each test case, output a line containing the largest possible area of a triangle built using three of the given sides (as a real number rounded to 2 decimal places). If it is not possible to construct any triangles then output "0.00" (quotes for clarity).

Sample input

2
4 3.0 4.0 5.0 100.0
3 1.0 2.0 4.0

Sample output

6.00
0.00

---

重新寫過了, 優化了許多部份,
答案的部分仍然是將邊做排序, 找連續三個可以湊成三角形的邊構成的最大三角形

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
inline double ReadPoint() {
    static char c;
    int p = 0;
    double t = 1;
    while((c = getchar()) >= '0')
        t /= 10, p = (p << 3) + (p << 1) + (c-'0');
    return (double)p*t;
}
inline int ReadDouble(double *x) {
    static char c, neg;
    while((c = getchar()) < '-')    {if(c == EOF) return EOF;}
    if(c == '.')    {*x = ReadPoint();return 0;}
    neg = (c == '-') ? -1 : 1;
    *x = (neg == 1) ? c-'0' : 0;
    while((c = getchar()) >= '0')
        *x = (*x)*10 + c-'0';
    if(c == '.')    *x += ReadPoint();
    *x *= neg;
    return 0;
}
int main() {
    int t, n, i;
    double A[10000];
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++) {
            ReadDouble(A+i);
        }
        sort(A, A+n);
        double area = 0.0, tmp, s;
        for(i = 2; i  < n; i++) {
            if(A[i-2]+A[i-1] <= A[i])
                continue;
            s = (A[i-2]+A[i-1]+A[i])/2;
            tmp = s*(s-A[i-2])*(s-A[i-1])*(s-A[i]);
            if(area < tmp)
                area = tmp;
        }
        area = sqrt(area);
        printf("%.2lf\n", area);
    }
    return 0;
}