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 }
