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('\n');

80         }

81     }

82     return 0;

83 }