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('\n');

101     }

102     return 0;

103 }