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。。
  1 //STATUS:C++_AC_190MS_1884KB
  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 long long LL;
 34 typedef unsigned long long ULL;
 35 //const
 36 const int N=100010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=100000,STA=8000010;
 39 //const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 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 stb[N<<1],ans[N];
 59 int T,n,x,y;
 60 
 61 int main(){
 62  //   freopen("in.txt","r",stdin);
 63     int i,j,k,ca=1,la,ra,lb,rb,reva,revb;
 64     scanf("%d",&T);
 65     while(T--)
 66     {
 67         scanf("%d%d%d",&n,&x,&y);
 68         k=reva=revb=0;
 69         la=1,ra=n;
 70         lb=rb=N-5;
 71         while(k<n){
 72             if(reva){
 73                 for(i=x;i-- && la<=ra;la++){
 74                     if(revb)stb[--lb]=la;
 75                     else stb[rb++]=la;
 76                 }
 77             }
 78             else {
 79                 for(i=x;i-- && la<=ra;ra--){
 80                     if(revb)stb[--lb]=ra;
 81                     else stb[rb++]=ra;
 82                 }
 83             }
 84             reva^=1;
 85             if(revb){
 86                 for(i=y;i-- && lb<rb;lb++)
 87                     ans[k++]=stb[lb];
 88             }
 89             else {
 90                 for(i=y;i-- && lb<rb;)
 91                     ans[k++]=stb[--rb];
 92             }
 93             revb^=1;
 94         }
 95 
 96         printf("Case %d:",ca++);
 97         for(i=n-1;i>=0;i--){
 98             printf(" %d",ans[i]);
 99         }
100         putchar('
');
101     }
102     return 0;
103 }