UVa 1042 Lots of Sunlight (枚举&最优斜率)

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1042 - Lots of Sunlight

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=3483

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<bits/stdc++.h>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;}