HDU-4627 The Unsolvable Problem 简单数学

  题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4627

  对n除个2,然后考虑下奇偶。。。

 1 //STATUS:C++_AC_15MS_228KB
 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 //using namespace __gnu_cxx;
25 //define
26 #define pii pair<int,int>
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define lson l,mid,rt<<1
29 #define rson mid+1,r,rt<<1|1
30 #define PI acos(-1.0)
31 //typedef
32 typedef __int64 LL;
33 typedef unsigned __int64 ULL;
34 //const
35 const int N=30;
36 const int INF=0x3f3f3f3f;
37 const int MOD=100000,STA=8000010;
38 const LL LNF=1LL<<60;
39 const double EPS=1e-8;
40 const double OO=1e15;
41 const int dx[4]={-1,0,1,0};
42 const int dy[4]={0,1,0,-1};
43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
44 //Daily Use ...
45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
55 //End
56 
57 LL T,n;
58 
59 int main()
60 {
61  //   freopen("in.txt","r",stdin);
62     LL i,j;
63     scanf("%I64d",&T);
64     while(T--)
65     {
66         scanf("%I64d",&n);
67         if(n&1)i=n/2,j=n/2+1;
68         else i=j=n/2;
69         while(gcd(i,j)!=1)i--,j++;
70         printf("%I64d\n",i*j);
71     }
72     return 0;
73 }

 

posted @ 2013-08-02 21:07  zhsl  阅读(189)  评论(0编辑  收藏  举报