UVa 1042 Lots of Sunlight (枚举&最优斜率) - c++编程基础

思路：对于给定的楼层，枚举左、右的最大斜率。

atan(k)/pi就表示所占时间的百分比，注意计算结束时间时要用pi-atan(k)处理下。

提交后居然返回了第一名。。

完整代码：

/*0.118s*/  
  
#include  
using namespace std;  
const double eps = 1e-8;  
const int BEGIN = (5 * 60 + 37) * 60, END = (18 * 60 + 17) * 60; /// 秒  
  
int m[105], d[105], pos[105];  
  
int main()  
{  
    int n, cas = 0, id, Floor, W, H, i, beg, end;  
    double left, right;  
    while (scanf("%d", &n), n)  
    {  
        scanf("%d%d", &W, &H);  
        for (i = 1; i < n; ++i)  
        {  
            scanf("%d%d", &m[i], &d[i]);  
            pos[i + 1] = pos[i] + W + d[i];  
        }  
        scanf("%d", &m[n]);  
        printf("Apartment Complex: %d\n", ++cas);  
        ///  
        while (scanf("%d", &Floor), Floor)  
        {  
            id = Floor % 100, Floor /= 100, left = right = 0.0;  
            if (id < 1 || id >




 n || Floor < 1 || Floor > m[id])  
            {  
                printf("Apartment %d: Does not exist\n", Floor * 100 + id);  
                continue;  
            }  
            for (i = 1; i < id; ++i)  
                if (m[i] >= Floor)  
                    left = max(left, (double)((m[i] - Floor + 1) * H) / (pos[id] - pos[i] - W));/// 高度差/距离  
            for (i = id + 1; i <= n; ++i)  
                if (m[i] >= Floor)  
                    right = max(right, (double)((m[i] - Floor + 1) * H) / (pos[i] - pos[id] - W));  
            beg = (int)(atan(left) * (END - BEGIN) / M_PI + eps) + BEGIN; ///注意+eps！  
            end = (int)((M_PI - atan(right)) * (END - BEGIN) / M_PI + eps) + BEGIN;  
            printf("Apartment %d: %02d:%02d:%02d - %02d:%02d:%02d\n", Floor * 100 + id,  
                   beg / 3600, beg / 60 % 60, beg % 60,  
                   end / 3600, end / 60 % 60, end % 60);  
        }  
    }  
    return 0;  
}