CF——Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)
2017-09-05 19:54
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C. Five Dimensional Points
1.2D时设原点为好点,所以最多周围有4个点,x,y正负轴上;3D时加2个(z轴),所以5D时最多周围由10个点,加上本身,最多11点。当n > 11时,输出0;
2.当n < 11 枚举,0 < t = cos <= 1 时是坏点;
AC code
1.2D时设原点为好点,所以最多周围有4个点,x,y正负轴上;3D时加2个(z轴),所以5D时最多周围由10个点,加上本身,最多11点。当n > 11时,输出0;
2.当n < 11 枚举,0 < t = cos <= 1 时是坏点;
AC code
#include <iostream> #include <string> #include<cstring> #include <stdio.h> #include <cmath> #include <algorithm> #include <vector> const int maxn = 1000 + 10; const int MOD = 1e9 + 7; const int INF = 0x3f3f3f3f; using namespace std; typedef long long LL; struct Node { int a; int b; int c; int d; int e; }T[maxn]; int ans[maxn],s=0; double dd(int i, int j) { return ((T[j].a-T[i].a)*(T[j].a - T[i].a) + (T[j].b-T[i].b)*(T[j].b-T[i].b)+(T[j].c-T[i].c)*(T[j].c- T[i].c)+(T[j].d-T[i].d)*(T[j].d-T[i].d) + (T[j].e-T[i].e)*(T[j].e-T[i].e)); } double dx(int i,int j,int k) { return ((T[j].a-T[i].a)*(T[k].a - T[i].a) + (T[j].b-T[i].b)*(T[k].b-T[i].b)+(T[j].c-T[i].c)*(T[k].c- T[i].c)+(T[j].d-T[i].d)*(T[k].d-T[i].d) + (T[j].e-T[i].e)*(T[k].e-T[i].e)); } int main() { int n; cin >> n; for(int i = 0; i < n; i++) scanf("%d%d%d%d%d",&T[i].a,&T[i].b,&T[i].c,&T[i].d,&T[i].e); if(n > 11) {cout << "0" << endl;return 0;} for(int i = 0; i < n; i++) { bool flag = false; for(int j = 0; j < n; j++) { if(j == i) continue; for(int k = 0; k < n; k++) { if(k == j || k == i) continue; double t = dx(i,j,k)/sqrt(dd(i,j)*dd(i,k)); if(t <= 1 && t > 0) flag = true; if(flag) break; } if(flag) break; } if(flag) continue; ans[s++] = i+1; } sort(ans,ans+s); cout << s << endl; for(int i = 0; i < s; i++) printf("%d\n",ans[i]); return 0; }
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