HDU-4671 Backup Plan 构造解

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4671

　　假设是3 m，首先按照第一列按照1 2 3 1 2 3 1...排下去，然后个数就是一个 (m/3)+1,(m/3)+1....m/3的形式，题目要求不相差1，那么对于第二列serve直接从后往前就可以了，总可以保证不想差1，最多两列就可以解决。。

 1 //STATUS:C++_AC_31MS_272KB
 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 LL MOD=1000000007,STA=8000010;
39 const LL LNF=1LL<<55;
40 const double EPS=1e-9;
41 const double OO=1e30;
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 ans[N];
59 int n,m;
60 
61 int main(){
62  //   freopen("in.txt","r",stdin);
63     int i,j,t,k;
64     while(~scanf("%d%d",&n,&m))
65     {
66         for(i=0;i<m && i<n;i++){
67             t=n-1;
68             for(j=i;j<m;j+=n,t=((t-1)+n)%n){
69                 if(t==i)t=((t-1)+n)%n;
70                 ans[j]=t;
71             }
72         }
73         for(i=0;i<m;i++){
74             printf("%d %d",i%n+1,ans[i]+1);
75             for(j=0;j<n;j++){
76                 if(j==i%n || j==ans[i])continue;
77                 printf(" %d",j+1);
78             }
79             putchar('
');
80         }
81     }
82     return 0;
83 }