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 }