UVa 1042 Lots of Sunlight: enumerations & Optimal slope

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

Idea: For a given floor, enumerate the maximum slope of the left and right.

Atan (k)/pi represents the percentage of the time taken, noting that the end time is calculated with Pi-atan (k).

I actually returned the first place after the submission.

Complete code:

/*0.118s*/#include <bits/stdc++.h> using namespace std;  
Const double EPS = 1e-8; const int BEGIN = (5 * + panax) *, end = (18 * 60 + 17) * 60;  
    
SEC 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%, Floor/=-MB, left = Righ  
            t = 0.0; if (ID < 1 | | | ID > N | | Floor < 1 | |  
                Floor > M[id]) {printf ("Apartment%d:does not exist\n", Floor * + ID);  
            Continue  }
            for (i = 1; i < ID; ++i) if (M[i] >= Floor) left = Max (left, D  
                ouble) ((M[i]-Floor + 1) * H)/(Pos[id]-pos[i]-W);///height difference/distance 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;  
            Pay attention to +eps!  
            end = (int) (M_pi-atan (right) * (End-begin)/M_pi + EPS) + BEGIN; printf ("Apartment%d:%02d:%02d:%02d-%02d:%02d:%02d\n", Floor * + ID, beg/3600, beg/60% 60  
        , Beg, end/3600, END/60, end% 60);  
} return 0; }

Author: csdn Blog Synapse7

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