A - Flip Game
总的情况数只有2^16次方种，显然直接bfs就可以了
1 #include<iostream> 2 #include<queue> 3 #include<cstring> 4 using namespace std; 5 int W,B,start; 6 bool have[1000000]; 7 struct plot{ 8 int n,step; 9 }; 10 void input(int &n){ 11 n=0; 12 char c; 13 for(int i=0;i<16;i++){ 14 cin>>c; 15 if(c=='b')n=n|(1<<i); 16 } 17 } 18 void turn(int &n,int k){ 19 n=n^(1<<k); 20 if(k%4)n=n^(1<<(k-1)); 21 if(k%4!=3)n=n^(1<<(k+1)); 22 if(k/4)n=n^(1<<(k-4)); 23 if(k/4<3)n=n^(1<<(k+4)); 24 } 25 bool bfs(){ 26 queue<plot> q; 27 plot p; 28 p.n=start;p.step=0; 29 q.push(p); 30 memset(have,0,sizeof(have)); 31 while(!q.empty()){ 32 int cur=q.front().n,step=q.front().step; 33 q.pop(); 34 if(have[cur])continue; 35 if(cur==W||cur==B){ 36 cout<<step<<endl; 37 return true; 38 } 39 for(int i=0;i<16;i++){ 40 p.n=cur; 41 turn(p.n,i); 42 p.step=step+1; 43 if(!have[p.n]) 44 q.push(p); 45 } 46 have[cur]=true; 47 } 48 return false; 49 } 50 int main(){ 51 W=0;B=0; 52 for(int i=0;i<16;i++){ 53 B=B|(1<<i); 54 } 55 input(start); 56 if(!bfs())cout<<"Impossible"<<endl; 57 return 0; 58 }
B - The Pilots Brothers' refrigerator
直接搜应该是可以的,但是不知道为什么一直T..
注意到如果要改变(i,j),只需要把这个十字上的全部点一遍,结果就只有(i,j)改变.改变了偶数次的相当于没有改变.
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 int have[4][4]; 5 int main(){ 6 for(int i=0;i<4;i++) 7 for(int j=0;j<4;j++)have[i][j]=0; 8 int a[4][4]; 9 char c; 10 for(int i=0;i<4;i++) 11 for(int j=0;j<4;j++){ 12 cin>>c; 13 if(c=='+')a[i][j]=0; 14 else a[i][j]=1; 15 } 16 for(int i=0;i<4;i++){ 17 for(int j=0;j<4;j++){ 18 if(a[i][j]==0){ 19 for(int k=0;k<4;k++){ 20 have[i][k]+=1; 21 have[k][j]+=1; 22 } 23 have[i][j]-=1; 24 } 25 } 26 } 27 int count=0; 28 for(int i=0;i<4;i++){ 29 for(int j=0;j<4;j++) 30 if(have[i][j]%2) 31 count++; 32 } 33 cout<<count<<endl; 34 for(int i=0;i<4;i++){ 35 for(int j=0;j<4;j++) 36 if(have[i][j]%2) 37 cout<<i+1<<' '<<j+1<<endl; 38 } 39 return 0; 40 }
C - Radar Installation
勾股定理先算一下每个小岛对应x轴上的一个区间,雷达放在这个区间内都可以扫到这个小岛,然后贪心一下就行了,按区间左端点排序,每次都放在未取点的区间左端点.
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 const int maxn=1005; 7 int n,d,x[maxn],y[maxn],k[maxn],Max[maxn]; 8 bool have[maxn]; 9 double l[maxn],r[maxn]; 10 bool Cmpk(int i,int j){ 11 if(l[i]==l[j])return r[i]>r[j]; 12 return l[i]>l[j]; 13 } 14 bool Cmpm(int i,int j){ 15 return r[i]>r[j]; 16 } 17 void turn(){ 18 for(int i=0;i<n;i++){ 19 l[i]=x[i]-sqrt((double)d*d-y[i]*y[i]); 20 r[i]=x[i]+sqrt((double)d*d-y[i]*y[i]); 21 } 22 sort(k,k+n,Cmpk); 23 sort(Max,Max+n,Cmpm); 24 } 25 int solve(){ 26 turn(); 27 int j=0,count=0; 28 for(int i=0;i<n;i++){ 29 if(have[k[i]])continue; 30 have[k[i]]=true; 31 count++; 32 while(j<n&&l[k[i]]<=r[Max[j]]){ 33 have[Max[j]]=true;j++; 34 } 35 } 36 return count; 37 } 38 int main(){ 39 int kase=0; 40 while(cin>>n>>d&&n){ 41 memset(have,0,sizeof(have)); 42 bool ans=true; 43 for(int i=0;i<n;i++){ 44 cin>>x[i]>>y[i]; 45 k[i]=i; 46 Max[i]=i; 47 if(y[i]>d){ 48 ans=false; 49 } 50 } 51 if(ans)cout<<"Case "<<++kase<<": "<<solve()<<endl; 52 else cout<<"Case "<<++kase<<": "<<-1<<endl; 53 } 54 return 0; 55 }
D - Power of Cryptography
显然要用高精度..然后二分一下答案..一开始一直T,应该是卡在二分的死循环上,如果一定有解,肯定可以跳出循环的..题目应该是有问题的,不一定有恰好等的k,而是k^n<=p,取最大的k,加上这个就过了...
1 #include<iostream> 2 #include<cstdio> 3 #include<string> 4 #include<vector> 5 #include<iomanip> 6 #include<cmath> 7 using namespace std; 8 const int BASE=1000000000,WIDTH=9; 9 typedef long long ll; 10 struct BigInt{ 11 bool neg; 12 vector<int> s; 13 BigInt operator =(string &str){ 14 s.clear(); 15 neg=false; 16 if(str[0]=='-'){ 17 neg=true; 18 str[0]='0'; 19 } 20 int x,len=(str.length()-1)/WIDTH+1; 21 for(int i=0;i<len;i++){ 22 int end=str.length()-i*WIDTH; 23 int start=max(0,end-WIDTH); 24 sscanf(str.substr(start,end-start).c_str(),"%d",&x); 25 s.push_back(x); 26 } 27 return *this; 28 } 29 BigInt operator =(int n){ 30 s.clear(); 31 if(n<0){ 32 neg=true; 33 s.push_back(-n); 34 }else{ 35 neg=false; 36 s.push_back(n); 37 } 38 return *this; 39 } 40 void print(){ 41 if(neg)cout<<'-'; 42 cout<<s.back(); 43 for(int i=s.size()-2;i>=0;i--){ 44 cout<<setw(8)<<setfill('0')<<s[i]; 45 } 46 cout<<endl; 47 } 48 }; 49 bool operator ==(const BigInt &a,const BigInt &b){ 50 if(a.neg!=b.neg)return false; 51 if(a.s.size()==b.s.size()){ 52 for(int i=0;i<a.s.size();i++) 53 if(a.s[i]!=b.s[i])return false; 54 return true; 55 } 56 return false; 57 } 58 bool operator<(BigInt &a,BigInt &b){ 59 if(a.neg&&!b.neg)return true; 60 if(!a.neg&&b.neg)return false; 61 if(a.neg&&b.neg){ 62 a.neg=b.neg=false; 63 bool ans=a<b||a==b; 64 a.neg=b.neg=true; 65 return !ans; 66 } 67 if(a.s.size()<b.s.size())return true; 68 else if(a.s.size()>b.s.size())return false; 69 for(int i=a.s.size()-1;i>=0;i--){ 70 if(a.s[i]==b.s[i])continue; 71 if(a.s[i]<b.s[i])return true; 72 return false; 73 } 74 return false; 75 } 76 bool operator >(BigInt &a,BigInt &b){return !(a==b&&a<b);} 77 bool operator <=(BigInt &a,BigInt &b){return (a==b||a<b);} 78 bool operator >=(BigInt &a,BigInt &b){return !(a<b);} 79 bool operator !=(BigInt &a,BigInt &b){return !(a==b);} 80 BigInt operator -(BigInt &a,BigInt &b); 81 BigInt operator +(BigInt &a,BigInt &b){ 82 BigInt ans; 83 if(a.neg&&!b.neg){ 84 a.neg=false; 85 ans=b-a; 86 a.neg=true; 87 return ans; 88 } 89 if(!a.neg&&b.neg){ 90 b.neg=false; 91 ans=a-b; 92 b.neg=true; 93 return ans; 94 } 95 ans.neg=a.neg; 96 int k=0,i=0; 97 while(i<a.s.size()||i<b.s.size()){ 98 if(i<a.s.size())k+=a.s[i]; 99 if(i<b.s.size())k+=b.s[i]; 100 ans.s.push_back(k%BASE); 101 k/=BASE; 102 i++; 103 } 104 if(k)ans.s.push_back(k); 105 return ans; 106 } 107 BigInt operator -(BigInt &a,BigInt &b){ 108 BigInt ans; 109 if(!a.neg&&b.neg){ 110 b.neg=false; 111 ans=a+b; 112 b.neg=true; 113 return ans; 114 } 115 if(a.neg&&!b.neg){ 116 a.neg=false; 117 ans=a+b; 118 a.neg=true; 119 ans.neg=true; 120 return ans; 121 } 122 if(a.neg&&b.neg&&a>b){ 123 b.neg=a.neg=false; 124 ans=b-a; 125 b.neg=a.neg=true; 126 return ans; 127 } 128 if(!a.neg&&!b.neg&&a<b){ 129 ans=b-a; 130 ans.neg=true; 131 return ans; 132 } 133 ans.neg=false; 134 int k=0,i=0; 135 while(i<a.s.size()||i<b.s.size()){ 136 if(i<a.s.size())k+=a.s[i]; 137 if(i<b.s.size())k-=b.s[i]; 138 if(k<0){ 139 ans.s.push_back(k+BASE); 140 k=-1; 141 }else{ 142 ans.s.push_back(k); 143 k=0; 144 } 145 i++; 146 } 147 while(ans.s.size()>1&&i>=0&&ans.s[--i]==0)ans.s.pop_back(); 148 return ans; 149 } 150 BigInt operator *(BigInt &a,BigInt &b){ 151 if(a.s.size()<b.s.size())return b*a; 152 BigInt ans; 153 ll k=0; 154 for(int l=0;l<a.s.size()+b.s.size()-1;l++){ 155 for(int i=min(l,int(a.s.size()-1));i>=0&&l-i<b.s.size();i--){ 156 k+=(ll)a.s[i]*b.s[l-i]; 157 } 158 ans.s.push_back(k%BASE); 159 k/=BASE; 160 } 161 if(k)ans.s.push_back(k); 162 ans.neg=a.neg^b.neg; 163 return ans; 164 } 165 BigInt operator *(BigInt &a,int b){ 166 BigInt ans; 167 ll k=0; 168 for(int l=0;l<a.s.size();l++){ 169 k+=a.s[l]*b; 170 ans.s.push_back(k%BASE); 171 k/=BASE; 172 } 173 if(k)ans.s.push_back(k); 174 if((b<0&&a.neg)||(b>0&&!a.neg))ans.neg=false; 175 else ans.neg=true; 176 return ans; 177 } 178 inline void devide_2(BigInt &a){ 179 int k=0; 180 // a.print(); 181 for(int i=a.s.size()-1;i>=0;i--){ 182 int k0=k; 183 k=a.s[i]%2==0?0:BASE/2; 184 //cout<<a.s[i]<<endl; 185 a.s[i]=k0+a.s[i]/2; 186 //cout<<a.s[i]<<endl; 187 } 188 if(a.s.back()==0)a.s.pop_back(); 189 } 190 inline BigInt devide(BigInt &a,BigInt &b,BigInt &r){ 191 BigInt L,R,m,t; 192 L=1;R=a; 193 if(a<b){ 194 r=a; 195 return m=0; 196 } 197 while(R>L){ 198 //L.print(); 199 //R.print(); 200 t=(R+L); 201 devide_2(t); 202 if(t==L){ 203 m=t*b; 204 r=a-m;return L; 205 } 206 m=t*b; 207 //m.print(); 208 if(m==a){ 209 r=0;return t; 210 }else if(m<a){ 211 L=t; 212 }else{ 213 R=t; 214 } 215 } 216 m=t*b; 217 r=a-m; 218 return L; 219 } 220 inline BigInt operator%(BigInt &a,BigInt &b){ 221 BigInt r; 222 devide(a,b,r); 223 return r; 224 } 225 inline bool Is_even(const BigInt &a){ 226 if(a.s[0]%2==0)return true; 227 return false; 228 } 229 //计算n的k次方模p 230 BigInt qpow(BigInt &n,int k){ 231 BigInt ans,y; 232 int _n=k; 233 //n.print();k.print();p.print(); 234 ans=1;y=n; 235 while(_n>=1){ 236 if(_n%2){ 237 ans=(ans*y); 238 //ans.print(); 239 ans=ans; 240 //ans.print(); 241 } 242 y=y*y; 243 //y.print(); 244 y=y; 245 //y.print(); 246 //cout<<"---------\n"; 247 _n/=2; 248 // _n.print(); 249 } 250 return ans; 251 } 252 BigInt p; 253 int n,len; 254 void solve(){ 255 BigInt l,r,m; 256 l=pow(10.0,(double)len/n-1); 257 r=pow(10.0,(double)len/n+1); 258 if(qpow(r,n)==p){ 259 r.print();return; 260 } 261 while(1){ 262 m=l+r; 263 devide_2(m); 264 if(m==l){ 265 l.print(); 266 return; 267 } 268 //m.print(); 269 BigInt q=qpow(m,n); 270 if(q==p){ 271 m.print();return; 272 } 273 if(q<p){ 274 l=m;continue; 275 }else{ 276 r=m;continue; 277 } 278 } 279 } 280 int main(){ 281 //freopen("e:\\in.txt","r",stdin); 282 string s; 283 while(cin>>n){ 284 cin>>s; 285 len=s.length(); 286 p=s; 287 solve(); 288 } 289 290 }
E - Y2K Accounting Bug
只有12个月,总的情况数是2^12次方,枚举一下.需要一些优化,比如目前可行解最多k个月盈利,那么小于等于k个月盈利的就可以不用算了.还可以通过s/d,比如s/d<=1/5,1/4<=s/d<=1/2都是不可能盈利的情况.
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 int m[12],s,d; 6 int surplus(){ 7 int ans=0; 8 int surmonth=0; 9 for(int i=0;i<(1<<12);i++){ 10 int count1=0; 11 for(int j=0;j<12;j++){ 12 count1+=(i&(1<<j))?1:0; 13 } 14 if(count1<=surmonth||count1*s-(12-count1)*d<=0)continue; 15 for(int j=0;j<12;j++) 16 m[j]=i&(1<<j); 17 bool ok=true; 18 for(int j=0;j<8;j++){ 19 int sur=0; 20 for(int k=0;k<5;k++){ 21 if(m[j+k])sur+=s; 22 else sur-=d; 23 } 24 if(sur>=0){ 25 ok=false;break; 26 } 27 } 28 if(ok){ 29 surmonth=count1; 30 int sur=count1*s-(12-count1)*d; 31 if(sur>ans)ans=sur; 32 } 33 } 34 return ans; 35 } 36 int main(){ 37 // freopen("e:\\in.txt","r",stdin); 38 // freopen("e:\\out.txt","w",stdout); 39 while(cin>>s>>d){ 40 // if((d<2*s&&3*s<2*d)||(d>4*s&&5*s>d))cout<<surplus()<<endl; 41 // else cout<<"Deficit"<<endl; 42 int ans=surplus(); 43 if(ans)cout<<ans<<endl; 44 else cout<<"Deficit"<<endl; 45 } 46 return 0; 47 }
F - Ubiquitous Religions
就是求连通块的个数,并查集求一下就行了..
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 const int maxn=50005; 6 int n,m,p[maxn]; 7 int Find(int x){return p[x]==x?x:p[x]=Find(p[x]);} 8 int main(){ 9 int kase=0; 10 while(scanf("%d%d",&n,&m)!=EOF&&n){ 11 int count=n; 12 for(int i=0;i<n;i++)p[i]=i; 13 int x,y; 14 for(int i=0;i<m;i++){ 15 scanf("%d%d",&x,&y); 16 x--;y--; 17 if(Find(x)==Find(y))continue; 18 p[Find(x)]=y; 19 count--; 20 } 21 printf("Case %d: %d\n",++kase,count); 22 } 23 return 0; 24 }
G - Tiling
显然可以得到递推式f(n)=f(n-1)+2*f(n-2),n只有250,也不需要矩阵了,高精度打个表...
1 import java.io.*; 2 import java.util.*; 3 import java.math.BigInteger; 4 public class Main { 5 public static void main(String arg[]){ 6 BigInteger a=BigInteger.valueOf(1),b=BigInteger.valueOf(1); 7 String s[]=new String[260]; 8 s[0]="1";s[1]="1"; 9 for(int i=2;i<=255;){ 10 a=b.add(a).add(a); 11 s[i++]=a.toString(); 12 b=a.add(b).add(b); 13 s[i++]=b.toString(); 14 } 15 Scanner sc=new Scanner(System.in); 16 int n; 17 while(sc.hasNext()){ 18 n=sc.nextInt(); 19 System.out.println(s[n]); 20 } 21 } 22 23 }
H - Tautology
只有5个小写字母,总的情况就是2^5,枚举一下是不是每次结果都为真就行了.式子的处理类似计算器,递归一下.
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 bool b[5]={0,0,0,0,0}; 5 int f(char c){ 6 switch(c){ 7 case 'p':return 0; 8 case 'q':return 1; 9 case 'r':return 2; 10 case 's':return 3; 11 case 't':return 4; 12 } 13 } 14 bool Is_up(char c){ 15 if(c>='A'&&c<='Z')return true; 16 return false; 17 } 18 bool Is_true(char *s){ 19 int count=0; 20 int n=strlen(s); 21 if(s[0]=='N')return !Is_true(s+1); 22 if(!Is_up(s[0]))return b[f(s[0])]; 23 int mid=1; 24 if(!Is_up(s[1]))mid=1; 25 else{ 26 mid=1; 27 for(int i=1;i<n;i++,mid++){ 28 if(s[i]=='N')count+=0; 29 else if(Is_up(s[i]))count+=1; 30 else count-=1; 31 if(count==-1)break; 32 } 33 } 34 switch(s[0]){ 35 case 'K':return Is_true(s+1)&&Is_true(s+mid+1); 36 case 'A':return Is_true(s+1)||Is_true(s+mid+1); 37 case 'C':return !(Is_true(s+1)==1&&Is_true(s+mid+1)==0); 38 case 'E':return Is_true(s+1)==Is_true(s+mid+1); 39 } 40 } 41 bool Is_tau(char *s){ 42 for(int i=0;i<32;i++){ 43 for(int j=0;j<5;j++) 44 b[j]=i&(1<<j); 45 if(!Is_true(s))return false; 46 } 47 return true; 48 } 49 int main(){ 50 char s[105]; 51 while(cin>>s&&s[0]!='0'){ 52 if(Is_tau(s))cout<<"tautology\n"; 53 else cout<<"not\n"; 54 } 55 return 0; 56 }
I - Parencodings
P先转S，S再转W
1 #include<iostream> 2 using namespace std; 3 const int maxn=25; 4 int s[2*maxn],p[maxn],w[maxn],n; 5 void p_s(){ 6 for(int i=n-1;i>0;i--) 7 p[i]-=p[i-1]; 8 int j=0; 9 for(int i=0;i<n;i++){ 10 while(p[i]){ 11 s[j++]=1; 12 p[i]--; 13 } 14 s[j++]=0; 15 } 16 } 17 void s_w(){ 18 int r[maxn],_n=0; 19 for(int i=0;i<2*n;i++){ 20 if(s[i]==0)r[_n++]=i; 21 } 22 for(int i=n-1;i>=0;i--){ 23 int j=r[i]-1,count=1; 24 while(j>=0&&count){ 25 if(s[j])count--; 26 else count++; 27 j--; 28 } 29 w[i]=(r[i]-j)/2; 30 } 31 } 32 int main(){ 33 int t; 34 cin>>t; 35 while(t--){ 36 cin>>n; 37 for(int i=0;i<n;i++) 38 cin>>p[i]; 39 p_s(); 40 s_w(); 41 for(int i=0;i<n;i++){ 42 if(i)cout<<' '; 43 cout<<w[i]; 44 } 45 cout<<endl; 46 } 47 return 0; 48 }
后面四道就是纯水模拟了...
J - Crashing Robots
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 const int maxn=105; 6 int house[maxn][maxn],A,B; 7 char d[4]={'N','E','S','W'}; 8 char ok[100]="OK"; 9 char ans[100]; 10 int f(char c){ 11 switch(c){ 12 case 'N':return 0; 13 case 'E':return 1; 14 case 'S':return 2; 15 case 'W':return 3; 16 } 17 } 18 //-1ÊÇǽ 19 int Is_crash(int x,int y){ 20 if(x==0||x==A+1||y==0||y==B+1)return -1; 21 return house[x][y]; 22 } 23 class robot{ 24 int x,y,order; 25 char dir; 26 public: 27 void input(int t){ 28 cin>>x>>y>>dir; 29 order=t; 30 house[x][y]=t; 31 } 32 void move(char c,int time){ 33 if(strcmp(ans,"OK"))return; 34 switch(c){ 35 case 'L':dir=d[(f(dir)-time%4+4)%4];break; 36 case 'R':dir=d[(f(dir)+time)%4];break; 37 default:switch(dir){ 38 case 'N': 39 for(int i=0;i<time;i++){ 40 y++; 41 int k=Is_crash(x,y); 42 if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;} 43 else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;} 44 house[x][y-1]=0; 45 house[x][y]=order; 46 } 47 break; 48 case 'S': 49 for(int i=0;i<time;i++){ 50 y--; 51 int k=Is_crash(x,y); 52 if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;} 53 else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;} 54 house[x][y+1]=0; 55 house[x][y]=order; 56 } 57 break; 58 case 'E': 59 for(int i=0;i<time;i++){ 60 x++; 61 int k=Is_crash(x,y); 62 if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;} 63 else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;} 64 house[x-1][y]=0; 65 house[x][y]=order; 66 } 67 break; 68 case 'W': 69 for(int i=0;i<time;i++){ 70 x--; 71 int k=Is_crash(x,y); 72 if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;} 73 else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;} 74 house[x+1][y]=0; 75 house[x][y]=order; 76 } 77 break; 78 } 79 } 80 } 81 }rb[maxn]; 82 int main(){ 83 //freopen("e:\\in.txt","r",stdin); 84 //freopen("e:\\out.txt","w",stdout); 85 int t,n,m,num,time; 86 char mo; 87 cin>>t; 88 while(t--){ 89 strcpy(ans,ok); 90 memset(house,0,sizeof(house)); 91 cin>>A>>B>>n>>m; 92 for(int i=1;i<=n;i++){ 93 rb[i].input(i); 94 } 95 for(int i=0;i<m;i++){ 96 cin>>num>>mo>>time; 97 rb[num].move(mo,time); 98 } 99 cout<<ans<<endl; 100 } 101 return 0; 102 }
K - Robot Motion
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 const int maxn=15; 6 char grid[maxn][maxn]; 7 int step[maxn][maxn],x,y,total,n,m,enter; 8 void solve(){ 9 memset(step,-1,sizeof(step)); 10 total=0;x=1;y=enter; 11 step[x][y]=0; 12 while(1){ 13 switch(grid[x][y]){ 14 case 'N':x--;break; 15 case 'S':x++;break; 16 case 'W':y--;break; 17 case 'E':y++;break; 18 } 19 total++; 20 if(x==0||x==n+1||y==0||y==m+1){ 21 cout<<total<<" step(s) to exit"<<endl;return; 22 } 23 if(step[x][y]!=-1){ 24 cout<<step[x][y]<<" step(s) before a loop of "<<total-step[x][y]<<" step(s)"<<endl;return; 25 } 26 step[x][y]=total; 27 } 28 } 29 int main(){ 30 while(cin>>n>>m>>enter&&n){ 31 for(int i=1;i<=n;i++){ 32 for(int j=1;j<=m;j++) 33 cin>>grid[i][j]; 34 } 35 solve(); 36 } 37 return 0; 38 }
L - Emag eht htiw Em Pleh
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 using namespace std; 5 char ans[20][35]; 6 bool Is_up(char c){ 7 if(c>='A'&&c<='Z')return true; 8 return false; 9 } 10 int main(){ 11 for(int i=0;i<=16;i+=2){ 12 strcpy(ans[i],"+---+---+---+---+---+---+---+---+"); 13 } 14 for(int i=3;i<=15;i+=4){ 15 strcpy(ans[i],"|:::|...|:::|...|:::|...|:::|...|"); 16 } 17 for(int i=1;i<=15;i+=4){ 18 strcpy(ans[i],"|...|:::|...|:::|...|:::|...|:::|"); 19 } 20 string s; 21 cin>>s; 22 cin>>s; 23 for(int i=0;i<s.size();){ 24 if(Is_up(s[i])){ 25 ans[17-2*(s[i+2]-'0')][4*(s[i+1]-'a')+2]=s[i]; 26 i+=4; 27 }else{ 28 ans[17-2*(s[i+1]-'0')][4*(s[i]-'a')+2]='P'; 29 i+=3; 30 } 31 } 32 cin>>s; 33 cin>>s; 34 for(int i=0;i<s.size();){ 35 if(Is_up(s[i])){ 36 ans[17-2*(s[i+2]-'0')][4*(s[i+1]-'a')+2]=s[i]-'A'+'a'; 37 i+=4; 38 }else{ 39 ans[17-2*(s[i+1]-'0')][4*(s[i]-'a')+2]='p'; 40 i+=3; 41 } 42 } 43 for(int i=0;i<=16;i++){ 44 cout<<ans[i]<<endl; 45 } 46 return 0; 47 }
M - Help Me with the Game
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<vector> 5 #include<algorithm> 6 using namespace std; 7 vector<int> w[26]; 8 vector<int> b[26]; 9 char order[]="KQRBN"; 10 bool Cmpw(int i,int j){ 11 if(i%16==j%16)return i/16<j/16; 12 return i%16<j%16; 13 } 14 bool Cmpb(int i,int j){ 15 if(i%16==j%16)return i/16<j/16; 16 return i%16>j%16; 17 } 18 bool Is_up(char c){ 19 if(c>='A'&&c<='Z')return true; 20 return false; 21 } 22 bool Is_low(char c){ 23 if(c>='a'&&c<='z')return true; 24 return false; 25 } 26 int main(){ 27 string s; 28 for(int i=0;i<26;i++){ 29 w[i].clear(); 30 b[i].clear(); 31 } 32 for(int i=8;i>=1;i--){ 33 getline(cin,s); 34 getline(cin,s); 35 for(int j=2;j<=30;j+=4){ 36 if(Is_up(s[j])){ 37 w[s[j]-'A'].push_back((j-2)*4+i); 38 }else if(Is_low(s[j])){ 39 b[s[j]-'a'].push_back((j-2)*4+i); 40 } 41 } 42 } 43 getline(cin,s); 44 cout<<"White: "; 45 for(int i=0;i<5;i++){ 46 char c=order[i]; 47 48 for(int j=0;j<w[c-'A'].size();j++){ 49 if(i||j)cout<<','; 50 cout<<c<<char(w[c-'A'][j]/16+'a')<<w[c-'A'][j]%16; 51 } 52 } 53 sort(w['P'-'A'].begin(),w['P'-'A'].end(),Cmpw); 54 for(int i=0;i<w['P'-'A'].size();i++){ 55 cout<<','<<char(w['P'-'A'][i]/16+'a')<<w['P'-'A'][i]%16; 56 } 57 cout<<endl; 58 cout<<"Black: "; 59 for(int i=0;i<5;i++){ 60 char c=order[i]; 61 for(int j=0;j<b[c-'A'].size();j++){ 62 if(i||j)cout<<','; 63 cout<<c<<char(b[c-'A'][j]/16+'a')<<b[c-'A'][j]%16; 64 } 65 } 66 sort(b['P'-'A'].begin(),b['P'-'A'].end(),Cmpb); 67 for(int i=0;i<b['P'-'A'].size();i++){ 68 cout<<','<<char(b['P'-'A'][i]/16+'a')<<b['P'-'A'][i]%16; 69 } 70 cout<<endl; 71 return 0; 72 }
