Title Source: http://codeforces.com/group/aUVPeyEnI2/contest/229510

Time limit: 2s

Space limit: 256MB

Main topic:

Given a convex polygon, there is a connection of two vertices to divide the polygon into two non-empty plots of an integer area, asking how many lines there are.

Data range:

4≤n≤200 000

? 109≤xi, yi≤109

Examples:

Code:

`#include <algorithm> #include <iostream> #include <cstring> #include <vector> #include < cstdio> #include <string> #include <cmath> #include <queue> #include <set> #include <map > #include <complex>using namespace std;typedef long long ll;typedef long double db;typedef pair<int,int> p Ii;typedef vector<int> VI; #define DE (x) cout << #x << "=" << x << endl#define Rep (i,a,b) for (int i=a;i< (b); ++i) #define ALL (x) (x). Begin (), (x). End () #define SZ (x) (int) (x). Size () #define MP Make_pair#define PB Push_back#define fi first#define se second#define pi acos ( -1.0) #define MEM0 (a) memset (A,0,sizeof (a)) #define MEMF (b) Memset (b,false,sizeof (b)) #define MAXN 401010int x[maxn],y[maxn];int va[maxn],pre[maxn],p[maxn][3][3][3];int across ( int A,int B,int C,int d) {return a*d-b*c;} int main () {freopen ("integral.in", "R", stdin); Freopen ("Integral.out", "w", stdout); MEM0 (P); MEM0 (pre); int n; Cin>>n; Rep (i,1,n+1) {cin>>x[i]>>y[i]; x[i]=x[i]&1; y[i]=y[i]&1; X[i+n]=x[i]; Y[i+n]=y[i]; } rep (i,1,n<<1|1) {Va[i]=across (x[i-1],y[i-1],x[i],y[i]); Pre[i]=pre[i-1]+va[i]; } if (pre[n+1]&1) {cout<< "0" <<endl; return 0; }//Rep (i,1,n<<1|1)//{//printf ("%5d", X[i]);//}//cout<<endl;//Rep (i,1,n<<1|1)//{// printf ("%5d", Y[i]);//}//cout<<endl;//Rep (i,1,n<<1|1)//{//printf ("%5d", Pre[i]);//}//cout<& Lt;endl; Rep (i,1,n<<1|1) Rep (a,0,2) Rep (b,0,2) Rep (c,0,2) {if (x[i]==a&&y[i]==b&& (pre[i]&1) ==c) p[i][a][b][c]=p[i-1][a][b][c]+1; else P[i][a][b][c]=p[i-1][a][b][c]; }//cout<<endl;//Rep (i,1,n+1)//REP (a,0,2)//REP (b,0,2)//REP (c,0,2)//{//Prin TF ("%5d", P[i][a][b][c]);//}//cout<<endl;//}cout<<endl; ll Ans=0; Rep (i,1,n+1) {int l=i+2,r=n+i-2; Rep (a,0,2) Rep (b,0,2) Rep (c,0,2) {int t= ((x[i]*b-y[i]*a+c-pre[i]) &1); if (t==0) ans+=p[r][a][b][c]-p[l-1][a][b][c];//printf ("%5d", ans); }//cout<<endl; } cout<<ans/2<<endl; return 0;}`

2016-2017 acm-icpc, NEERC, Northern subregional Contest problem I. Integral polygons