BNUOJ-29357 Bread Sorting 模拟

　　题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=29357

　　直接模拟就可以了。。

 //STATUS:C++_AC_190MS_1884KB
 #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 long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 //const LL LNF=1LL<<60;
 const double EPS=1e-;
 const double OO=1e15;
 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 stb[N<<],ans[N];
 int T,n,x,y;

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,k,ca=,la,ra,lb,rb,reva,revb;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d%d",&n,&x,&y);
         k=reva=revb=;
         la=,ra=n;
         lb=rb=N-;
         while(k<n){
             if(reva){
                 for(i=x;i-- && la<=ra;la++){
                     if(revb)stb[--lb]=la;
                     else stb[rb++]=la;
                 }
             }
             else {
                 for(i=x;i-- && la<=ra;ra--){
                     if(revb)stb[--lb]=ra;
                     else stb[rb++]=ra;
                 }
             }
             reva^=;
             if(revb){
                 for(i=y;i-- && lb<rb;lb++)
                     ans[k++]=stb[lb];
             }
             else {
                 for(i=y;i-- && lb<rb;)
                     ans[k++]=stb[--rb];
             }
             revb^=;
         }

         printf("Case %d:",ca++);
         for(i=n-;i>=;i--){
             printf(" %d",ans[i]);
         }
         putchar('\n');
     }
     return ;
 }