Marisa’s Cake
Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 210 Accepted Submission(s): 123
Marisa wants to know the expected size of the cake she might get, and because you are madly in love with her, you decided to do anything she wants for her! You take out your laptop and are ready to calculate the expected size for Marisa. However, Marisa is bad with fractions. She wants to make sure the answer she gets is always an integer. To do that, she would like you to multiply the answer with the total number of possible cakes there are. Unfortunately, that still doesn’t guarantee the answer be an integer. An additional 2 must also be multiplied into the answer as well. For example, let
2 4 0 0 1 0 1 1 0 2 5 1 1 3 1 3 2 2 3 1 2
9 50solution:给定一个多边形的顶点位置,问你由这些顶点构成的多边形的面积是多少面积乘
2 的话直接算叉积就好了,例如三个点A,B,C 构成的三角形,设从原点到他们的向量为fA,fB,fC ,那么三角形面积的2 倍就是fA×fB+fB×fC+fC×fA ,同理k 个点(对应向量按逆时针排序为f1,f2,...,fk )构成的多边形面积的2 倍等于∑k−1i=1fi×fi+1+fk×f1 。
直接算每种情况的多边形求面积是O(n2n) 的,但是我们可以发现计算面积的过程中有很多的重复计算。
对于两个在原多边形上的点i,j ,我们可以算出它们产生的贡献fi×fj 在多少个需要算的多边形M={a0,a1,...,ap−1} 的式子里出现。
首先如果at=i ,那么必须有a(t+1)modp=j ,也就是说按逆时针顺序在i 和j 之间的点都不能选,而且为了保证p≥3 ,所以按逆时针顺序在j 到i 这一段的点至少要有一个在多边形M 上,设j 到i 之间有c 个点,那么对应的方案数就是2c−1 ,于是我们得到了任意两个点之间向量叉积的系数。
于是答案可以被表示为∑i=1n∑j=1i−1(2i−j−1−1)(fi×fj)+(2n−i+j−1−1)(fj×fi)=∑i=1n(2i−1fi)×(∑j=1i−12−jfj)+(∑j=1i−12jfj)×(2n−i−1fi)
如果令
g(x)=∑i=1x2ifih(x)=∑i=1x2−ifi
答案就是
∑i=1n(2i−1fi)×h(i−1)+g(i−1)×(2n−i−1fi)
于是在模意义下预处理2i 和2−i ,边利用前缀和的特点维护g(i),h(i) ,边统计答案就行了,时间复杂度O(n) 。#include<cstdio> using namespace std; const int mod = 1e9 + 7; const int maxn = 100050; int n ; long long mul[maxn], inv[maxn]; struct Point{ int x, y; Point(int x = 0, int y = 0) :x(x), y(y){} int operator *(const Point &p){ return (1ll * x*p.y%mod - 1ll * y*p.x%mod + mod) % mod; } Point operator *(const long long p){ return Point(p*x%mod, y*p%mod); } Point operator + (const Point &p){ return Point((x + p.x) % mod, (y + p.y) % mod); } }e,mul1,inv1; int main() { int t; scanf("%d", &t); int inv2 = 500000004;//由费马小定理求出 mul[0] = inv[0] = 1; for (int i = 1; i < maxn; i++) { mul[i] = (2 * mul[i - 1]) % mod; inv[i] = (inv2*inv[i - 1]) % mod; }//求2^i 和2^(-i)的前缀 while (t--) { long long ans = 0; scanf("%d", &n); mul1 = inv1 = 0; for (int i = 1; i <= n; i++) { scanf("%d%d", &e.x, &e.y); ans = (ans + (e*mul[i - 1])* inv1) % mod; if (i == n)ans = (ans + mul1*(e*inv2)) % mod;//此时n-i-1为-1 因此要用2的逆元inv2 else ans = (ans + mul1*(e*mul[n - i - 1])) % mod; mul1 = mul1 + e*mul[i]; inv1 = inv1 + e*inv[i]; } printf("%I64d\n", ans); } return 0; }