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题意：有n个正方形和一个角(均在第一象限中)，使这些正方形与这个角构成封闭的阴影区域，求阴影区域面积的最大值。
析：很容易知道只有所有的正方形的对角形在一条直线时，是最大的，然后根据数学关系，就容易得到答案。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define print(a) printf("%d\n", (a))
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int main(){
double sq, s, ax, ay, bx, by;
while(scanf("%d", &n) && n){
cin >> ax >> ay >> bx >> by;
sq = 0.0, s = 0.0;
for(int i = 0; i < n; i++){
double m; cin >> m;
s += m * m;
sq += m;
}
double k1 = ay / ax, k2 = by / bx;
double x1 = fabs((k2+1)*sq/(k2-k1)), y1 = k1*x1;
double x2 = fabs((k1+1)*sq/(k2-k1)), y2 = k2*x2;
double area = fabs((x1*y2 - x2*y1) / 2.0);
double sum = area - s/2.0;
printf("%.3lf\n", sum);
}
return 0;
}
UVa 1643 Angle and Squares (计算几何)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5910240.html
