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。。

 //STATUS:C++_AC_31MS_272KB
 #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=;
 const int INF=0x3f3f3f3f;
 //const LL MOD=1000000007,STA=8000010;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e30;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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 ans[N];
 int n,m;

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,t,k;
     while(~scanf("%d%d",&n,&m))
     {
         for(i=;i<m && i<n;i++){
             t=n-;
             for(j=i;j<m;j+=n,t=((t-)+n)%n){
                 if(t==i)t=((t-)+n)%n;
                 ans[j]=t;
             }
         }
         for(i=;i<m;i++){
             printf("%d %d",i%n+,ans[i]+);
             for(j=;j<n;j++){
                 if(j==i%n || j==ans[i])continue;
                 printf(" %d",j+);
             }
             putchar('\n');
         }
     }
     return ;
 }