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=110;
const int INF=0x3f3f3f3f;
//const LL MOD=1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e30;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=0;i<m && i<n;i++){
            t=n-1;
            for(j=i;j<m;j+=n,t=((t-1)+n)%n){
                if(t==i)t=((t-1)+n)%n;
                ans[j]=t;
            }
        }
        for(i=0;i<m;i++){
            printf("%d %d",i%n+1,ans[i]+1);
            for(j=0;j<n;j++){
                if(j==i%n || j==ans[i])continue;
                printf(" %d",j+1);
            }
            putchar('
');
        }
    }
    return 0;
}