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题意:给一个凸多边形,求任选若干点形成的多边形的面积和。
思路:
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#pragma comment(linker, "/STACK:10240000") #include <bits/stdc++.h> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; namespace Debug { void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<" ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} } template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} /* -------------------------------------------------------------------------------- */ const int maxn = 1e5 + 7; const int mod = 1e9 + 7; struct Point { ll x, y; void read() { int x, y; scanf("%d%d", &x, &y); this->x = x; this->y = y; } Point(int x, int y) { this->x = x; this->y = y; } Point() {} Point operator + (const Point &that) const { return Point((x + that.x) % mod, (y + that.y) % mod); } Point operator * (const int &m) const { return Point(x * m % mod, y * m % mod); } }; Point p[maxn], g[maxn], h[maxn]; int two[maxn], twoinv[maxn]; int cross(Point &a, Point &b) { return (a.x * b.y % mod - a.y * b.x % mod + mod) % mod; } int powermod(int a, int n, int mod) { ll ans = 1, buf = a; while (n) { if (n & 1) ans = (ans * buf) % mod; buf = buf * buf % mod; n >>= 1; } return ans; } void init() { ll inv = powermod(2, mod - 2, mod); two[0] = twoinv[0] = 1; for (int i = 1; i < maxn; i ++) { two[i] = (two[i - 1] + two[i - 1]) % mod; twoinv[i] = twoinv[i - 1] * inv % mod; } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, n; cin >> T; init(); while (T --) { cin >> n; for (int i = 1; i <= n; i ++) { p[i].read(); g[i] = g[i - 1] + p[i] * two[i - 1]; h[i] = h[i - 1] + p[i] * twoinv[i + 1]; } int ans = 0; for (int i = 2; i <= n; i ++) { Point buf = p[i] * two[n - i]; ans = (ans + cross(g[i - 1], buf)) % mod; } for (int i = 2; i <= n; i ++) { Point buf = p[i] * two[i]; ans = (ans + cross(buf, h[i - 1])) % mod; } cout << ans << endl; } } |
[hdu5448 Marisa’s Cake]多边形面积,公式化简
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原文地址:http://www.cnblogs.com/jklongint/p/4809042.html