URAL-1981 Parallel and Perpendicular 水题
2013-10-30 00:14
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题目链接:/article/6701620.html
水题,注意细节。
水题,注意细节。
//STATUS:C++_AC_31MS_333KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream> #include <iomanip> #include <numeric> #include <cstring> #include <cassert> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <list> #include <set> #include <map> using namespace std; //#pragma comment(linker,"/STACK:102400000,102400000") //using namespace __gnu_cxx; //define #define pii pair<int,int> #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define PI acos(-1.0) //typedef typedef __int64 LL; typedef unsigned __int64 ULL; //const const int N=110; const int INF=0x3f3f3f3f; const int MOD=95041567,STA=8000010; const LL LNF=1LL<<60; const double EPS=1e-8; const double OO=1e15; const int dx[4]={-1,0,1,0}; const int dy[4]={0,1,0,-1}; const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; //Daily Use ... inline int sign(double x){return (x>EPS)-(x<-EPS);} template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} template<class T> inline T lcm(T a,T b,T d){return a/d*b;} template<class T> inline T Min(T a,T b){return a<b?a:b;} template<class T> inline T Max(T a,T b){return a>b?a:b;} template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} //End int n; int main() { // freopen("in.txt","r",stdin); int i,j; LL ans1,ans2; while(~scanf("%d",&n)) { if(n==4){ printf("0 2\n"); continue; } if(n==5){ printf("0 0\n"); continue; } if(n==6){ printf("6 9\n"); continue; } if(n&1){ ans1=(LL)n*(n-3)/2; printf("%I64d 0\n",ans1); } else { ans1=(LL)n*(n-3)/2; printf("%I64d %I64d\n",ans1,ans1); } } return 0; }
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