hzwer已经说的很好了，在此只能跪烂了
1 /************************************************************** 2 Problem: 1914 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:88 ms 7 Memory:3160 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 14 using namespace std; 15 typedef long long ll; 16 typedef double lf; 17 const int N = 100005; 18 19 int n; 20 ll cnt = 0; 21 22 struct P { 23 ll x, y; 24 lf an; 25 P() {} 26 P(ll _x, ll _y, lf _an) : x(_x), y(_y), an(_an) {} 27 }a[N]; 28 inline bool operator < (const P &a, const P &b) { 29 return a.an < b.an; 30 } 31 inline ll operator * (const P &a, const P &b) { 32 return (ll) a.x * b.y - a.y * b.x; 33 } 34 35 inline int read() { 36 int x = 0, sgn = 1; 37 char ch = getchar(); 38 while (ch < '0' || '9' < ch) { 39 if (ch == '-') sgn = -1; 40 ch = getchar(); 41 } 42 while ('0' <= ch && ch <= '9') { 43 x = x * 10 + ch - '0'; 44 ch = getchar(); 45 } 46 return sgn * x; 47 } 48 49 void work() { 50 int r = 1, t = 0, i; 51 for (i = 1; i <= n; ++i) { 52 while ((r % n + 1) != i && a[i] * a[r % n + 1] >= 0) ++t, ++r; 53 cnt += (ll) t * (t - 1) / 2; 54 --t; 55 } 56 } 57 58 int main() { 59 n = read(); 60 int i, X, Y; 61 for (i = 1; i <= n; ++i) { 62 X = read(), Y = read(); 63 a[i] = P(X, Y, atan2(Y, X)); 64 } 65 sort(a + 1, a + n + 1); 66 work(); 67 printf("%lld ", (ll) n * (n - 1) * (n - 2) / 6 - cnt); 68 return 0; 69 }
话说为了Rank 1我又丧病的使用了fread...还正是神器啊Orz
1 /************************************************************** 2 Problem: 1914 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:60 ms 7 Memory:4728 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 14 using namespace std; 15 typedef long long ll; 16 typedef double lf; 17 const int N = 100005; 18 const int Maxbuf = 1600005; 19 int n; 20 ll cnt = 0, Left = 0, len; 21 char buf[Maxbuf]; 22 23 struct P { 24 ll x, y; 25 lf an; 26 P() {} 27 P(ll _x, ll _y, lf _an) : x(_x), y(_y), an(_an) {} 28 }a[N]; 29 inline bool operator < (const P &a, const P &b) { 30 return a.an < b.an; 31 } 32 inline ll operator * (const P &a, const P &b) { 33 return (ll) a.x * b.y - a.y * b.x; 34 } 35 36 inline int read() { 37 int x = 0, sgn = 1; 38 while (buf[Left] < '0' || '9' < buf[Left]) { 39 if (buf[Left] == '-') sgn = -1; 40 ++Left; 41 } 42 while ('0' <= buf[Left] && buf[Left] <= '9') { 43 x = x * 10 + buf[Left] - '0'; 44 ++Left; 45 } 46 return sgn * x; 47 } 48 49 void work() { 50 int r = 1, t = 0, i; 51 for (i = 1; i <= n; ++i) { 52 while ((r % n + 1) != i && a[i] * a[r % n + 1] >= 0) ++t, ++r; 53 cnt += (ll) t * (t - 1) / 2; 54 --t; 55 } 56 } 57 58 int main() { 59 len = fread(buf, 1, Maxbuf, stdin); 60 buf[len] = ' '; 61 n = read(); 62 int i, X, Y; 63 for (i = 1; i <= n; ++i) { 64 X = read(), Y = read(); 65 a[i] = P(X, Y, atan2(Y, X)); 66 } 67 sort(a + 1, a + n + 1); 68 work(); 69 printf("%lld ", (ll) n * (n - 1) * (n - 2) / 6 - cnt); 70 return 0; 71 }
（p.s. 比Rank 2快了2ms 23333）