NKU 专题一 题解

A - Flip Game

总的情况数只有2^16次方种，显然直接bfs就可以了

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int W,B,start;
bool have[1000000];
struct plot{
    int n,step;
};
void input(int &n){
    n=0;
    char c;
    for(int i=0;i<16;i++){
        cin>>c;
        if(c=='b')n=n|(1<<i);
    }
}
void turn(int &n,int k){
    n=n^(1<<k);
    if(k%4)n=n^(1<<(k-1));
    if(k%4!=3)n=n^(1<<(k+1));
    if(k/4)n=n^(1<<(k-4));
    if(k/4<3)n=n^(1<<(k+4));
}
bool bfs(){
    queue<plot> q;
    plot p;
    p.n=start;p.step=0;
    q.push(p);
    memset(have,0,sizeof(have));
    while(!q.empty()){
        int cur=q.front().n,step=q.front().step;
        q.pop();
        if(have[cur])continue;
        if(cur==W||cur==B){
            cout<<step<<endl;
            return true;
        }
        for(int i=0;i<16;i++){
            p.n=cur;
            turn(p.n,i);
            p.step=step+1;
            if(!have[p.n])
                q.push(p);
        }
        have[cur]=true;
    }
    return false;
}
int main(){
    W=0;B=0;
    for(int i=0;i<16;i++){
        B=B|(1<<i);
    }
    input(start);
    if(!bfs())cout<<"Impossible"<<endl;
    return 0;
}

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.

#include<iostream>
#include<cstring>
using namespace std;
int have[4][4];
int main(){
    for(int i=0;i<4;i++)
        for(int j=0;j<4;j++)have[i][j]=0;
    int a[4][4];
    char c;
    for(int i=0;i<4;i++)
    for(int j=0;j<4;j++){
        cin>>c;
        if(c=='+')a[i][j]=0;
        else a[i][j]=1;
    }
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++){
            if(a[i][j]==0){
                for(int k=0;k<4;k++){
                    have[i][k]+=1;
                    have[k][j]+=1;
                }
                have[i][j]-=1;
            }
        }
    }
    int count=0;
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++)
            if(have[i][j]%2)
                count++;
    }
    cout<<count<<endl;
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++)
            if(have[i][j]%2)
                cout<<i+1<<' '<<j+1<<endl;
    }
    return 0;
}

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.

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=1005;
int n,d,x[maxn],y[maxn],k[maxn],Max[maxn];
bool have[maxn];
double l[maxn],r[maxn];
bool Cmpk(int i,int j){
    if(l[i]==l[j])return r[i]>r[j];
    return l[i]>l[j];
}
bool Cmpm(int i,int j){
    return r[i]>r[j];
}
void turn(){
    for(int i=0;i<n;i++){
        l[i]=x[i]-sqrt((double)d*d-y[i]*y[i]);
        r[i]=x[i]+sqrt((double)d*d-y[i]*y[i]);
    }
    sort(k,k+n,Cmpk);
    sort(Max,Max+n,Cmpm);
}
int solve(){
    turn();
    int j=0,count=0;
    for(int i=0;i<n;i++){
        if(have[k[i]])continue;
        have[k[i]]=true;
        count++;
        while(j<n&&l[k[i]]<=r[Max[j]]){
            have[Max[j]]=true;j++;
        }
    }
    return count;
}
int main(){
    int kase=0;
    while(cin>>n>>d&&n){
        memset(have,0,sizeof(have));
        bool ans=true;
        for(int i=0;i<n;i++){
            cin>>x[i]>>y[i];
            k[i]=i;
            Max[i]=i;
            if(y[i]>d){
                ans=false;
            }
        }
        if(ans)cout<<"Case "<<++kase<<": "<<solve()<<endl;
        else cout<<"Case "<<++kase<<": "<<-1<<endl;
    }
    return 0;
}

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...

#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<iomanip>
#include<cmath>
using namespace std;
const int BASE=1000000000,WIDTH=9;
typedef long long ll;
struct BigInt{
    bool neg;
    vector<int> s;
    BigInt operator =(string &str){
        s.clear();
        neg=false;
        if(str[0]=='-'){
            neg=true;
            str[0]='0';
        }
        int x,len=(str.length()-1)/WIDTH+1;
        for(int i=0;i<len;i++){
            int end=str.length()-i*WIDTH;
            int start=max(0,end-WIDTH);
            sscanf(str.substr(start,end-start).c_str(),"%d",&x);
            s.push_back(x);
        }
        return *this;
    }
    BigInt operator =(int n){
        s.clear();
        if(n<0){
            neg=true;
            s.push_back(-n);
        }else{
            neg=false;
            s.push_back(n);
        }
        return *this;
    }
    void print(){
        if(neg)cout<<'-';
        cout<<s.back();
        for(int i=s.size()-2;i>=0;i--){
            cout<<setw(8)<<setfill('0')<<s[i];
        }
        cout<<endl;
    }
};
bool operator ==(const BigInt &a,const BigInt &b){
    if(a.neg!=b.neg)return false;
    if(a.s.size()==b.s.size()){
        for(int i=0;i<a.s.size();i++)
            if(a.s[i]!=b.s[i])return false;
        return true;
    }
    return false;
}
bool operator<(BigInt &a,BigInt &b){
    if(a.neg&&!b.neg)return true;
    if(!a.neg&&b.neg)return false;
    if(a.neg&&b.neg){
        a.neg=b.neg=false;
        bool ans=a<b||a==b;
        a.neg=b.neg=true;
        return !ans;
    }
    if(a.s.size()<b.s.size())return true;
    else if(a.s.size()>b.s.size())return false;
    for(int i=a.s.size()-1;i>=0;i--){
        if(a.s[i]==b.s[i])continue;
        if(a.s[i]<b.s[i])return true;
        return false;
    }
    return false;
}
bool operator >(BigInt &a,BigInt &b){return !(a==b&&a<b);}
bool operator <=(BigInt &a,BigInt &b){return (a==b||a<b);}
bool operator >=(BigInt &a,BigInt &b){return !(a<b);}
bool operator !=(BigInt &a,BigInt &b){return !(a==b);}
BigInt operator -(BigInt &a,BigInt &b);
BigInt operator +(BigInt &a,BigInt &b){
    BigInt ans;
    if(a.neg&&!b.neg){
        a.neg=false;
        ans=b-a;
        a.neg=true;
        return ans;
    }
    if(!a.neg&&b.neg){
        b.neg=false;
        ans=a-b;
        b.neg=true;
        return ans;
    }
    ans.neg=a.neg;
    int k=0,i=0;
    while(i<a.s.size()||i<b.s.size()){
        if(i<a.s.size())k+=a.s[i];
        if(i<b.s.size())k+=b.s[i];
        ans.s.push_back(k%BASE);
        k/=BASE;
        i++;
    }
    if(k)ans.s.push_back(k);
    return ans;
}
BigInt operator -(BigInt &a,BigInt &b){
    BigInt ans;
    if(!a.neg&&b.neg){
        b.neg=false;
        ans=a+b;
        b.neg=true;
        return ans;
    }
    if(a.neg&&!b.neg){
        a.neg=false;
        ans=a+b;
        a.neg=true;
        ans.neg=true;
        return ans;
    }
    if(a.neg&&b.neg&&a>b){
        b.neg=a.neg=false;
        ans=b-a;
        b.neg=a.neg=true;
        return ans;
    }
    if(!a.neg&&!b.neg&&a<b){
        ans=b-a;
        ans.neg=true;
        return ans;
    }
    ans.neg=false;
    int k=0,i=0;
    while(i<a.s.size()||i<b.s.size()){
        if(i<a.s.size())k+=a.s[i];
        if(i<b.s.size())k-=b.s[i];
        if(k<0){
            ans.s.push_back(k+BASE);
            k=-1;
        }else{
            ans.s.push_back(k);
            k=0;
        }
        i++;
    }
    while(ans.s.size()>1&&i>=0&&ans.s[--i]==0)ans.s.pop_back();
    return ans;
}
BigInt operator *(BigInt &a,BigInt &b){
    if(a.s.size()<b.s.size())return b*a;
    BigInt ans;
    ll k=0;
    for(int l=0;l<a.s.size()+b.s.size()-1;l++){
        for(int i=min(l,int(a.s.size()-1));i>=0&&l-i<b.s.size();i--){
            k+=(ll)a.s[i]*b.s[l-i];
        }
        ans.s.push_back(k%BASE);
        k/=BASE;
    }
    if(k)ans.s.push_back(k);
    ans.neg=a.neg^b.neg;
    return ans;
}
BigInt operator *(BigInt &a,int b){
    BigInt ans;
    ll k=0;
    for(int l=0;l<a.s.size();l++){
        k+=a.s[l]*b;
        ans.s.push_back(k%BASE);
        k/=BASE;
    }
    if(k)ans.s.push_back(k);
    if((b<0&&a.neg)||(b>0&&!a.neg))ans.neg=false;
    else ans.neg=true;
    return ans;
}
inline void devide_2(BigInt &a){
    int k=0;
   // a.print();
    for(int i=a.s.size()-1;i>=0;i--){
        int k0=k;
        k=a.s[i]%2==0?0:BASE/2;
        //cout<<a.s[i]<<endl;
        a.s[i]=k0+a.s[i]/2;
        //cout<<a.s[i]<<endl;
    }
    if(a.s.back()==0)a.s.pop_back();
}
inline BigInt devide(BigInt &a,BigInt &b,BigInt &r){
    BigInt L,R,m,t;
    L=1;R=a;
    if(a<b){
        r=a;
        return m=0;
    }
    while(R>L){
        //L.print();
        //R.print();
        t=(R+L);
        devide_2(t);
        if(t==L){
            m=t*b;
            r=a-m;return L;
        }
        m=t*b;
        //m.print();
        if(m==a){
            r=0;return t;
        }else if(m<a){
            L=t;
        }else{
            R=t;
        }
    }
    m=t*b;
    r=a-m;
    return L;
}
inline BigInt operator%(BigInt &a,BigInt &b){
    BigInt r;
    devide(a,b,r);
    return r;
}
inline bool Is_even(const BigInt &a){
    if(a.s[0]%2==0)return true;
    return false;
}
//计算n的k次方模p
BigInt qpow(BigInt &n,int k){
    BigInt ans,y;
    int _n=k;
    //n.print();k.print();p.print();
    ans=1;y=n;
    while(_n>=1){
        if(_n%2){
            ans=(ans*y);
            //ans.print();
            ans=ans;
            //ans.print();
        }
        y=y*y;
        //y.print();
        y=y;
        //y.print();
        //cout<<"---------
";
        _n/=2;
       // _n.print();
    }
    return ans;
}
BigInt p;
int n,len;
void solve(){
    BigInt l,r,m;
    l=pow(10.0,(double)len/n-1);
    r=pow(10.0,(double)len/n+1);
    if(qpow(r,n)==p){
        r.print();return;
    }
    while(1){
        m=l+r;
        devide_2(m);
        if(m==l){
            l.print();
            return;
        }
        //m.print();
        BigInt q=qpow(m,n);
        if(q==p){
            m.print();return;
        }
        if(q<p){
            l=m;continue;
        }else{
            r=m;continue;
        }
    }
}
int main(){
    //freopen("e:\in.txt","r",stdin);
    string s;
    while(cin>>n){
        cin>>s;
        len=s.length();
        p=s;
        solve();
    }

}

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.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int m[12],s,d;
int surplus(){
    int ans=0;
    int surmonth=0;
    for(int i=0;i<(1<<12);i++){
        int count1=0;
        for(int j=0;j<12;j++){
            count1+=(i&(1<<j))?1:0;
        }
        if(count1<=surmonth||count1*s-(12-count1)*d<=0)continue;
        for(int j=0;j<12;j++)
            m[j]=i&(1<<j);
            bool ok=true;
        for(int j=0;j<8;j++){
            int sur=0;
            for(int k=0;k<5;k++){
                if(m[j+k])sur+=s;
                else sur-=d;
            }
            if(sur>=0){
                ok=false;break;
            }
        }
        if(ok){
            surmonth=count1;
            int sur=count1*s-(12-count1)*d;
            if(sur>ans)ans=sur;
        }
    }
    return ans;
}
int main(){
//    freopen("e:\in.txt","r",stdin);
//    freopen("e:\out.txt","w",stdout);
    while(cin>>s>>d){
//        if((d<2*s&&3*s<2*d)||(d>4*s&&5*s>d))cout<<surplus()<<endl;
//        else cout<<"Deficit"<<endl;
        int ans=surplus();
        if(ans)cout<<ans<<endl;
        else cout<<"Deficit"<<endl;
    }
    return 0;
}

F - Ubiquitous Religions

就是求连通块的个数,并查集求一下就行了..

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=50005;
int n,m,p[maxn];
int Find(int x){return p[x]==x?x:p[x]=Find(p[x]);}
int main(){
    int kase=0;
    while(scanf("%d%d",&n,&m)!=EOF&&n){
        int count=n;
        for(int i=0;i<n;i++)p[i]=i;
        int x,y;
        for(int i=0;i<m;i++){
            scanf("%d%d",&x,&y);
            x--;y--;
            if(Find(x)==Find(y))continue;
            p[Find(x)]=y;
            count--;
        }
        printf("Case %d: %d
",++kase,count);
    }
    return 0;
}

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...

import java.io.*;
import java.util.*;
import java.math.BigInteger;
public class Main {
    public static void main(String arg[]){
        BigInteger a=BigInteger.valueOf(1),b=BigInteger.valueOf(1);
        String s[]=new String[260];
        s[0]="1";s[1]="1";
        for(int i=2;i<=255;){
            a=b.add(a).add(a);
            s[i++]=a.toString();
            b=a.add(b).add(b);
            s[i++]=b.toString();
        }
        Scanner sc=new Scanner(System.in);
        int n;
        while(sc.hasNext()){
            n=sc.nextInt();
            System.out.println(s[n]);
        }
    }

}

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.

#include<iostream>
#include<cstring>
using namespace std;
bool b[5]={0,0,0,0,0};
int f(char c){
    switch(c){
    case 'p':return 0;
    case 'q':return 1;
    case 'r':return 2;
    case 's':return 3;
    case 't':return 4;
    }
}
bool Is_up(char c){
    if(c>='A'&&c<='Z')return true;
    return false;
}
bool Is_true(char *s){
    int count=0;
    int n=strlen(s);
    if(s[0]=='N')return !Is_true(s+1);
    if(!Is_up(s[0]))return b[f(s[0])];
    int mid=1;
    if(!Is_up(s[1]))mid=1;
    else{
        mid=1;
        for(int i=1;i<n;i++,mid++){
            if(s[i]=='N')count+=0;
            else if(Is_up(s[i]))count+=1;
            else count-=1;
            if(count==-1)break;
        }
    }
    switch(s[0]){
        case 'K':return Is_true(s+1)&&Is_true(s+mid+1);
        case 'A':return Is_true(s+1)||Is_true(s+mid+1);
        case 'C':return !(Is_true(s+1)==1&&Is_true(s+mid+1)==0);
        case 'E':return Is_true(s+1)==Is_true(s+mid+1);
    }
}
bool Is_tau(char *s){
    for(int i=0;i<32;i++){
        for(int j=0;j<5;j++)
            b[j]=i&(1<<j);
        if(!Is_true(s))return false;
    }
    return true;
}
int main(){
    char s[105];
    while(cin>>s&&s[0]!='0'){
        if(Is_tau(s))cout<<"tautology
";
        else cout<<"not
";
    }
    return 0;
}

I - Parencodings

 P先转S，S再转W

#include<iostream>
using namespace std;
const int maxn=25;
int s[2*maxn],p[maxn],w[maxn],n;
void p_s(){
    for(int i=n-1;i>0;i--)
        p[i]-=p[i-1];
    int j=0;
    for(int i=0;i<n;i++){
        while(p[i]){
            s[j++]=1;
            p[i]--;
        }
        s[j++]=0;
    }
}
void s_w(){
    int r[maxn],_n=0;
    for(int i=0;i<2*n;i++){
        if(s[i]==0)r[_n++]=i;
    }
    for(int i=n-1;i>=0;i--){
        int j=r[i]-1,count=1;
        while(j>=0&&count){
            if(s[j])count--;
            else count++;
            j--;
        }
        w[i]=(r[i]-j)/2;
    }
}
int main(){
    int t;
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>p[i];
        p_s();
        s_w();
        for(int i=0;i<n;i++){
            if(i)cout<<' ';
            cout<<w[i];
        }
        cout<<endl;
    }
    return 0;
}

后面四道就是纯水模拟了...

J - Crashing Robots

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=105;
int house[maxn][maxn],A,B;
char d[4]={'N','E','S','W'};
char ok[100]="OK";
char ans[100];
int f(char c){
    switch(c){
        case 'N':return 0;
        case 'E':return 1;
        case 'S':return 2;
        case 'W':return 3;
    }
}
//-1ÊÇǽ
int Is_crash(int x,int y){
    if(x==0||x==A+1||y==0||y==B+1)return -1;
    return house[x][y];
}
class robot{
    int x,y,order;
    char dir;
public:
    void input(int t){
        cin>>x>>y>>dir;
        order=t;
        house[x][y]=t;
    }
    void move(char c,int time){
        if(strcmp(ans,"OK"))return;
        switch(c){
            case 'L':dir=d[(f(dir)-time%4+4)%4];break;
            case 'R':dir=d[(f(dir)+time)%4];break;
            default:switch(dir){
            case 'N':
                for(int i=0;i<time;i++){
                    y++;
                    int k=Is_crash(x,y);
                    if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;}
                    else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;}
                    house[x][y-1]=0;
                    house[x][y]=order;
                }
                break;
            case 'S':
                for(int i=0;i<time;i++){
                    y--;
                    int k=Is_crash(x,y);
                    if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;}
                    else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;}
                    house[x][y+1]=0;
                    house[x][y]=order;
                }
                break;
            case 'E':
                for(int i=0;i<time;i++){
                    x++;
                    int k=Is_crash(x,y);
                    if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;}
                    else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;}
                    house[x-1][y]=0;
                    house[x][y]=order;
                }
                break;
            case 'W':
                for(int i=0;i<time;i++){
                    x--;
                    int k=Is_crash(x,y);
                    if(k==-1){sprintf(ans,"Robot %d crashes into the wall",order);break;}
                    else if(k){sprintf(ans,"Robot %d crashes into robot %d",order,k);break;}
                    house[x+1][y]=0;
                    house[x][y]=order;
                }
                break;
            }
        }
    }
}rb[maxn];
int main(){
    //freopen("e:\in.txt","r",stdin);
    //freopen("e:\out.txt","w",stdout);
    int t,n,m,num,time;
    char mo;
    cin>>t;
    while(t--){
        strcpy(ans,ok);
        memset(house,0,sizeof(house));
        cin>>A>>B>>n>>m;
        for(int i=1;i<=n;i++){
            rb[i].input(i);
        }
        for(int i=0;i<m;i++){
            cin>>num>>mo>>time;
            rb[num].move(mo,time);
        }
        cout<<ans<<endl;
    }
    return 0;
}

K - Robot Motion

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=15;
char grid[maxn][maxn];
int step[maxn][maxn],x,y,total,n,m,enter;
void solve(){
    memset(step,-1,sizeof(step));
    total=0;x=1;y=enter;
    step[x][y]=0;
    while(1){
        switch(grid[x][y]){
            case 'N':x--;break;
            case 'S':x++;break;
            case 'W':y--;break;
            case 'E':y++;break;
        }
        total++;
        if(x==0||x==n+1||y==0||y==m+1){
            cout<<total<<" step(s) to exit"<<endl;return;
        }
        if(step[x][y]!=-1){
            cout<<step[x][y]<<" step(s) before a loop of "<<total-step[x][y]<<" step(s)"<<endl;return;
        }
        step[x][y]=total;
    }
}
int main(){
    while(cin>>n>>m>>enter&&n){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++)
                cin>>grid[i][j];
        }
        solve();
    }
    return 0;
}

L - Emag eht htiw Em Pleh

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
char ans[20][35];
bool Is_up(char c){
    if(c>='A'&&c<='Z')return true;
    return false;
}
int main(){
    for(int i=0;i<=16;i+=2){
        strcpy(ans[i],"+---+---+---+---+---+---+---+---+");
    }
    for(int i=3;i<=15;i+=4){
        strcpy(ans[i],"|:::|...|:::|...|:::|...|:::|...|");
    }
    for(int i=1;i<=15;i+=4){
        strcpy(ans[i],"|...|:::|...|:::|...|:::|...|:::|");
    }
    string s;
    cin>>s;
    cin>>s;
    for(int i=0;i<s.size();){
        if(Is_up(s[i])){
            ans[17-2*(s[i+2]-'0')][4*(s[i+1]-'a')+2]=s[i];
            i+=4;
        }else{
            ans[17-2*(s[i+1]-'0')][4*(s[i]-'a')+2]='P';
            i+=3;
        }
    }
    cin>>s;
    cin>>s;
    for(int i=0;i<s.size();){
        if(Is_up(s[i])){
            ans[17-2*(s[i+2]-'0')][4*(s[i+1]-'a')+2]=s[i]-'A'+'a';
            i+=4;
        }else{
            ans[17-2*(s[i+1]-'0')][4*(s[i]-'a')+2]='p';
            i+=3;
        }
    }
    for(int i=0;i<=16;i++){
        cout<<ans[i]<<endl;
    }
    return 0;
}

M - Help Me with the Game

#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> w[26];
vector<int> b[26];
char order[]="KQRBN";
bool Cmpw(int i,int j){
    if(i%16==j%16)return i/16<j/16;
    return i%16<j%16;
}
bool Cmpb(int i,int j){
    if(i%16==j%16)return i/16<j/16;
    return i%16>j%16;
}
bool Is_up(char c){
    if(c>='A'&&c<='Z')return true;
    return false;
}
bool Is_low(char c){
    if(c>='a'&&c<='z')return true;
    return false;
}
int main(){
    string s;
    for(int i=0;i<26;i++){
        w[i].clear();
        b[i].clear();
    }
    for(int i=8;i>=1;i--){
        getline(cin,s);
        getline(cin,s);
        for(int j=2;j<=30;j+=4){
            if(Is_up(s[j])){
                w[s[j]-'A'].push_back((j-2)*4+i);
            }else if(Is_low(s[j])){
                b[s[j]-'a'].push_back((j-2)*4+i);
            }
        }
    }
    getline(cin,s);
    cout<<"White: ";
    for(int i=0;i<5;i++){
        char c=order[i];

        for(int j=0;j<w[c-'A'].size();j++){
            if(i||j)cout<<',';
            cout<<c<<char(w[c-'A'][j]/16+'a')<<w[c-'A'][j]%16;
        }
    }
    sort(w['P'-'A'].begin(),w['P'-'A'].end(),Cmpw);
    for(int i=0;i<w['P'-'A'].size();i++){
        cout<<','<<char(w['P'-'A'][i]/16+'a')<<w['P'-'A'][i]%16;
    }
    cout<<endl;
    cout<<"Black: ";
    for(int i=0;i<5;i++){
        char c=order[i];
        for(int j=0;j<b[c-'A'].size();j++){
            if(i||j)cout<<',';
            cout<<c<<char(b[c-'A'][j]/16+'a')<<b[c-'A'][j]%16;
        }
    }
    sort(b['P'-'A'].begin(),b['P'-'A'].end(),Cmpb);
    for(int i=0;i<b['P'-'A'].size();i++){
        cout<<','<<char(b['P'-'A'][i]/16+'a')<<b['P'-'A'][i]%16;
    }
    cout<<endl;
    return 0;
}