URAL-1981 Parallel and Perpendicular 水题
题目链接：http://www.cnblogs.com/zhsl/p/3395868.html
水题，注意细节。
1 //STATUS:C++_AC_31MS_333KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef __int64 LL; 34 typedef unsigned __int64 ULL; 35 //const 36 const int N=110; 37 const int INF=0x3f3f3f3f; 38 const int MOD=95041567,STA=8000010; 39 const LL LNF=1LL<<60; 40 const double EPS=1e-8; 41 const double OO=1e15; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 int n; 59 60 int main() 61 { 62 // freopen("in.txt","r",stdin); 63 int i,j; 64 LL ans1,ans2; 65 while(~scanf("%d",&n)) 66 { 67 if(n==4){ 68 printf("0 2\n"); 69 continue; 70 } 71 if(n==5){ 72 printf("0 0\n"); 73 continue; 74 } 75 if(n==6){ 76 printf("6 9\n"); 77 continue; 78 } 79 if(n&1){ 80 ans1=(LL)n*(n-3)/2; 81 printf("%I64d 0\n",ans1); 82 } 83 else { 84 ans1=(LL)n*(n-3)/2; 85 printf("%I64d %I64d\n",ans1,ans1); 86 } 87 } 88 return 0; 89 }
