狗狗40题~（Volume B）

H - Sorting Slides

应该是个二分匹配的模板题的，但我还不会写 = =

其实数据规模很小，就用贪心的方法就水过了（没加vis判冲突wa了几发，从此开始艰难的没有1A 的生活）

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 200000000
 typedef long long ll;
 struct slide{
    int x1,x2,y1,y2;
 }rec[];
 int g[][],d[],ans[],vis[];
 int main(){
 //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
    int n,cs=;
    while(scanf("%d",&n) && n){
       for(int i=;i<n;i++)scanf("%d%d%d%d",&rec[i].x1,&rec[i].x2,&rec[i].y1,&rec[i].y2);
       int x,y;
       memset(d,,sizeof d);
       memset(g,,sizeof g);
       memset(vis,,sizeof vis);
       memset(ans,-,sizeof ans);
       for(int i=;i<n;i++){
          scanf("%d%d",&x,&y);
          for(int j=;j<n;j++)if(x<rec[j].x2&&x>rec[j].x1&&y<rec[j].y2&&y>rec[j].y1){
              g[i][j+n]=;d[i]++;
              d[j+n]++;
          }
       }
       int flag=;
       while(flag){
          int i,j,i0,j0;
          flag=;
          for(i=;i<n;i++)if(!vis[i]&&d[i]==)
             {flag=;break;}
          if(flag){
             for(j=n;j<*n;j++)if(!vis[j]&&g[i][j])break;
             i0=i,j0=j;
             ans[j0-n]=i0;
             vis[i0]=vis[j0]=;
             g[i0][j0]=;d[i0]--;d[j0]--;
             for(i=;i<n;i++)if(g[i][j0])d[i]--,g[i][j0]=;
             continue;
          }
          for(j=n;j<*n;j++)if(!vis[j]&&d[j]==)
             {flag=;break;}
          if(flag){
             for(i=;i<n;i++)if(!vis[i]&&g[i][j])break;
             i0=i;j0=j;
             ans[j0-n]=i0;
             vis[i0]=vis[j0]=;
             g[i0][j0]=;d[i0]--;d[j0]--;
             for(j=n;j<*n;j++)if(g[i0][j])d[j]--,g[i0][j]=;
             continue;
          }
       }
       int cnt=;
       printf("Heap %d\n",cs++);
       for(int i=;i<n;i++)if(ans[i]>=)
          {if(cnt)printf(" ");printf("(%c,%d)",'A'+i,ans[i]+);cnt++;}
       if(!cnt)printf("none");
       printf("\n\n");
       }
    return ;
 }

A - Triangle Encapsulation

水题，暴搜の

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 200000000

 typedef long long ll;
 struct point {
     int x,y;
     struct point operator -(const struct point b){
         struct point c;
         c.x=x-b.x;c.y=y-b.y;
         return c;
     }
 }p[];
 int cmp(struct point a,struct point b){
    if(a.y-b.y)return a.y>b.y;
    return a.x<b.x;
 }
 int g[][];
 int cross(struct point a,struct point b){
    return a.x*b.y-a.y*b.x;
 }
 int main(){
 //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
     printf("Program 4 by team X\n");
    while(scanf("%d%d",&p[].x,&p[].y)!=EOF){
       for(int i=;i<;i++)scanf("%d%d",&p[i].x,&p[i].y);
       struct point pi;
       memset(g,,sizeof g);
       int xmin=,ymin=,ymax=,xx[]={};
       for(pi.x=-;pi.x<=;pi.x++)
        for(pi.y=-;pi.y<=;pi.y++){
            int c1=,c2=;
            for(int i=;i<;i++){
               if(cross(pi-p[i],pi-p[(i+)%])<)c1++;
               if(cross(pi-p[i],pi-p[(i+)%])>)c2++;
            }
            if(c1==||c2==){
               g[pi.x+][pi.y+]=;
               xmin=min(xmin,pi.x+);
               xx[pi.y+]=max(xx[pi.y+],pi.x+);
               ymin=min(ymin,pi.y+);
               ymax=max(ymax,pi.y+);
            }
        }
       int x,y;
       for(y=ymax;y>=ymin;y--){
         for(x=xmin;x<=xx[y];x++){
            if(x>xmin)printf(" ");
            if(g[x][y])printf("(%2d,%3d)",x-,y-);
            else printf("        ");
         }
         printf("\n");
       }
       printf("\n");
    }
    printf("End of program 4 by team X\n");
    return ;
 }

G - Reflections

几何题。解析算错很多次，其实就是算交点和关于直线对称。用向量点积和叉积不容易错。

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>
 #include <utility>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 200000000
 #define eps 1e-6

 typedef long long ll;
 int n,ans[],cur;
 struct point{
    double x,y;
 };
 struct circle{
    double x,y,r;
 }cir[];
 struct line{
    double x0,y0,dx,dy;
 }ori;
 int sgn(double x){
    if(x>eps)return ;
    if(x<-eps)return -;
    return ;
 }
 pair <double,int> inter(struct line now){
    double a,b,c,del,t1,t2,t=-;
    int use=-;
    for(int i=;i<=n;i++){
       a=now.dx*now.dx+now.dy*now.dy;
       b=now.dx*(now.x0-cir[i].x)+now.dy*(now.y0-cir[i].y);
       c=(now.x0-cir[i].x)*(now.x0-cir[i].x)+(now.y0-cir[i].y)*(now.y0-cir[i].y)-cir[i].r*cir[i].r;
       del=b*b-a*c;
       if(sgn(del)<=)continue;
       t1=(-b-sqrt(del))/a;
       t2=(-b+sqrt(del))/a;
       if(sgn(t1)>){
          if(t==-||t>t1)t=t1,use=i;
       }
       else if(sgn(t2)>){
          if(t==-||t>t2)t=t2,use=i;
       }
    }
    return make_pair(t,use);
 }
 void work(struct line now){
    if(cur>)return;
    double t=inter(now).first;
    if(t<=)return;
    int i0=ans[cur++]=inter(now).second;

    /* 求对称射线 */
    double x,y,xx,yy;
    x=now.x0+t*now.dx;
    y=now.y0+t*now.dy;
    if(sgn(now.dy*(x-cir[i0].x)-now.dx*(y-cir[i0].y))==){
        now.x0=x;now.y0=y;
        now.dx=-now.dx;now.dy=-now.dy;
        work(now);
        return;
    }
    xx=-((x-cir[i0].x)*now.dx+(y-cir[i0].y)*now.dy)/cir[i0].r;
    yy=((x-cir[i0].x)*now.dy-(y-cir[i0].y)*now.dx)/cir[i0].r;
    now.x0=x;now.y0=y;
    now.dx=(xx*(x-cir[i0].x)-yy*(y-cir[i0].y))/cir[i0].r;
    now.dy=(xx*(y-cir[i0].y)+yy*(x-cir[i0].x))/cir[i0].r;
    work(now);
 }      

 int main(){
 //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
    int cs=;
    while(scanf("%d",&n) && n){
        int i;
        for(i=;i<=n;i++)scanf("%lf%lf%lf",&cir[i].x,&cir[i].y,&cir[i].r);
        scanf("%lf%lf%lf%lf",&ori.x0,&ori.y0,&ori.dx,&ori.dy);
        cur=;
        work(ori);
        printf("Scene %d\n",cs++);
        for(i=;i<cur&&i<;i++)printf("%d ",ans[i]);
        if(cur>)printf("...\n\n");
        else printf("inf\n\n");
    }
    return ;
 }

C - Robbery

一开始有点不敢写，想了一下其实每个时间点只和前后两个点有关，所以正着跑一次再倒着check一遍就确定了。

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 200000000

 typedef long long ll;
 int w,h,T,n;
 int g[][][];
 struct message{
    int t,l,u,r,d;
 }mes[];
 struct answer{
    int t,x,y;
 }ans[];
 int cmp(struct message a,struct message b){
    return a.t<b.t;
 }
 void direct(){
    int k=;
       memset(g,-,sizeof g);
       memset(g[],,sizeof g[]);
       for(int t=;t<=T;t++){
         //to expand reachable area.
         if(t>)
         for(int i=;i<=h;i++)
            for(int j=;j<=w;j++)if(!g[t-][i][j]){
                  g[t][i][j]=;
                  if(i>)g[t][i-][j]=;
                  if(j>)g[t][i][j-]=;
                  if(i<h)g[t][i+][j]=;
                  if(j<w)g[t][i][j+]=;
            }
         //to receive the messge.
         for(;k<n&&mes[k].t==t;k++){
            for(int i=mes[k].u;i<=mes[k].d;i++)
              for(int j=mes[k].l;j<=mes[k].r;j++)
                 g[t][i][j]=-;
         }

       }
 }
 void inverse(){
    int k=n-;
    for(int t=T-;t>;t--){
       for(int i=;i<=h;i++)
         for(int j=;j<=w;j++)if(!g[t][i][j]){
             if(!g[t+][i][j])continue;
             if(i>&&!g[t+][i-][j])continue;
             if(j>&&!g[t+][i][j-])continue;
             if(i<h&&!g[t+][i+][j])continue;
             if(j<w&&!g[t+][i][j+])continue;
             g[t][i][j]=-;
         }
    }
 }
 int main(){
     //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
    int cs=;
    while(scanf("%d%d%d",&w,&h,&T) && w){
       scanf("%d",&n);
       for(int i=;i<n;i++)
          scanf("%d%d%d%d%d",&mes[i].t,&mes[i].l,&mes[i].u,&mes[i].r,&mes[i].d);
       sort(mes,mes+n,cmp);
       direct();
       inverse();
       printf("Robbery #%d:\n",cs++);
       int res=,flag=;
       for(int t=;t<=T;t++){
         int cn=,x,y;
         for(int i=;i<=h;i++)
           for(int j=;j<=w;j++)if(!g[t][i][j])x=i,y=j,cn++;
         if(!cn){flag=;break;}
         if(cn==)ans[res].t=t,ans[res].x=x,ans[res++].y=y;
       }
       if(!flag)printf("The robber has escaped.\n\n");
       else{
          if(!res)printf("Nothing known.\n\n");
          else{
             for(int i=;i<res;i++)printf("Time step %d: The robber has been at %d,%d.\n",ans[i].t,ans[i].y,ans[i].x);
             printf("\n");
          }
       }
    }
    return ;
 }

E - Triangle War

状压dp，极大极小算法。

另外有个坑点是每次把能取的三角形位置都取完不一定是最优策略。。

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 200000000

 typedef long long ll;
 int edge[][],tri[]={,,,,,,,,};
 int dp[<<];
 int check(int sta){
    int ans=;
    for(int i=;i<;i++)if((sta&tri[i])==tri[i])ans++;
    return ans;
 }
 void dfs(){
    int s=(<<)-;
    dp[s]=;
    for(s=s-;s>=;s--){
       int ori=check(s),cn,maxdp=-;
       for(int i=;i<;i++)if(!((<<i)&s)){
          int to=(<<i)|s;
          int temp=check(to)-ori;
          if(temp)temp+=dp[to];
          else temp-=dp[to];
          maxdp=max(maxdp,temp);
       }
       dp[s]=maxdp;
    }
 }
 int main(){
 //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
    int t,gm,G;
    edge[][]=;edge[][]=<<;
    edge[][]=<<;edge[][]=<<;edge[][]=<<;
    edge[][]=<<;edge[][]=<<;
    edge[][]=<<;edge[][]=<<;edge[][]=<<;
    edge[][]=<<;edge[][]=<<;edge[][]=<<;
    edge[][]=<<;edge[][]=<<;
    edge[][]=<<;edge[][]=<<;edge[][]=<<;
    dfs();
    scanf("%d",&t);
    while(t--){
       scanf("%d",&G);
       for(gm=;gm<=G;gm++){
          int m,sta=,u,v,turn=;
          int a[]={};
          scanf("%d",&m);
          for(int i=;i<=m;i++){
             scanf("%d%d",&u,&v);
             int ori=check(sta);
             sta+=edge[u][v];
             ori=check(sta)-ori;
             a[turn]+=ori;
             if(!ori)turn=!turn;
          }
          printf("Game %d: ",gm);
          a[turn]+=dp[sta];
          if(a[]>a[])printf("A wins.\n");
          else printf("B wins.\n");
       }
       if(t)printf("\n");
    }
    return ;
 }

B - Symbolic Derivation

可以用string数组代替栈来做模拟

 #include <stdio.h>
 #include <cstring>
 #include <iostream>
 #include <cstdlib>
 #include <stack>
 #include <cctype>
 #include <string>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 2000000000

 typedef long long ll;
 char sgn[];
 string dev[];
 string oii[];
 string add(string a,string b,char op){
    if(op=='+'){
       string temp=a;
       if(b[]!='+'&&b[]!='-')temp+="+";
       temp+=b;
       return temp;
    }else{
       string temp=a;
       if(b[]=='+')b.replace(,,"-");
       else if(b[]=='-')b.replace(,,"+");
       else temp+="-";
       temp+=b;
       return temp;
    }
 }
 void multi(int &po,int &pd){
    string o2=oii[--po];
    string o1=oii[--po];
    string d2=dev[--pd];
    string d1=dev[--pd];
       string t1="(";
       t1+=d1;t1+="*";t1+=o2;
       string t2;
       t2+=o1;t2+="*";t2+=d2;t2+=")";
       dev[pd++]=add(t1,t2,'+');
       o1+="*";o1+=o2;
       oii[po++]=o1;
 }
 void divi(int &po,int &pd){
    string o2=oii[--po];
    string o1=oii[--po];
    string d2=dev[--pd];
    string d1=dev[--pd];
    string t1="(";
    t1+=d1;t1+="*";t1+=o2;
    string t2;
    t2+=o1;t2+="*";t2+=d2;t2+=")/";
    string temp=add(t1,t2,'-');
    temp+=o2;temp+="^2";
    dev[pd++]=temp;
    o1+="/";o1+=o2;
    oii[po++]=o1;
 }
 void loge(int &po,int &pd){
    string o=oii[--po];
    string d=dev[--pd];
             string temp="ln";
             temp+=o;
             oii[po++]=temp;
             temp="";
             temp+=d;temp+="/";
             temp+=o;
             dev[pd++]=temp;
 }
 string derivation(char *s,int n){
    int sg=,dv=,oi=;
    bool beg=;
    for(int i=;i<n;i++){
       if(!beg&&(s[i]=='+'||s[i]=='-')){
          char *p=s+i;
          if(s[++i]!='x'){
            while(isdigit(s[i])||s[i]=='.')i++;
            string temp=string(p,s+i);
            oii[oi++]=temp;
            dev[dv++]="";
          }
          else{
            string temp=s[i]=='-'?"-x":"+x";
            oii[oi++]=temp;
            temp=s[i]=='-'?"-1":"+1";
            dev[dv++]=temp;
            ++i;
          }
       }
       beg=;
       if(isdigit(s[i])){
          char *p=s+i;
          while(isdigit(s[i])||s[i]=='.')i++;
          string temp=string(p,s+i);
          oii[oi++]=temp;
          dev[dv++]="";
       }
       if(s[i]=='x'){
          oii[oi++]="x";
          dev[dv++]="";
          ++i;
       }
       if(s[i]=='l'){
          i++;
          sgn[sg++]='l';
       }
       else if(s[i]=='*'||s[i]=='/'){
          if(sg&&(sgn[sg-]=='*'||sgn[sg-]=='/')){
            char u=sgn[--sg];
            if(u=='*'){
               multi(oi,dv);
            }
            if(u=='/'){
               divi(oi,dv);
            }
          }
          sgn[sg++]=s[i];
       }
       else if(s[i]=='+'||s[i]=='-'){
          if(sg&&sgn[sg-]!='('){
            char u=sgn[--sg];
            if(u=='*'){
               multi(oi,dv);
            }
            if(u=='/'){
               divi(oi,dv);
            }
            if(u=='+'||u=='-'){
               string o1=oii[--oi];
               oii[oi-]=add(oii[oi-],o1,u);
               string d1=dev[--dv];
               dev[dv-]=add(dev[dv-],d1,u);
            }
          }
          sgn[sg++]=s[i];
       }
       else if(s[i]=='(')sgn[sg++]=s[i],beg=;
       else if(s[i]==')'){
          char u;
          while(sg&&sgn[sg-]!='('){
            u=sgn[--sg];
            if(u=='*'){
               multi(oi,dv);
            }
            if(u=='/'){
               divi(oi,dv);
            }
            if(u=='+'||u=='-'){
               string o1=oii[--oi];
               oii[oi-]=add(oii[oi-],o1,u);
               string d1=dev[--dv];
               dev[dv-]=add(dev[dv-],d1,u);
            }
          }
          //add brackets.
          if(u!='*'&&u!='/'){
             string temp="(";
             temp+=dev[--dv];
             temp+=")";
             dev[dv++]=temp;
             temp="(";
             temp+=oii[--oi];
             temp+=")";
             oii[oi++]=temp;
          }
          sg--;
          if(sgn[sg-]=='l'){
             loge(oi,dv);
             sg--;
          }
       }
    }
    char u;
    while(sg){
      u=sgn[--sg];
      if(u=='*'){
      multi(oi,dv);
      }
      if(u=='/'){
      divi(oi,dv);
      }
      if(u=='+'||u=='-'){
      string o1=oii[--oi];
      oii[oi-]=add(oii[oi-],o1,u);
      string d1=dev[--dv];
      dev[dv-]=add(dev[dv-],d1,u);
      }
    }
    return dev[dv-];
 }

 char buf[];
 int main(){
    //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
    while(~scanf("%s",buf)){
       string ans=derivation(buf,strlen(buf));
       cout << ans << endl;
    }
    return ;
 }

F - Metal Cutting

半平面交，中间思考人生把排序写挫了好久才看出来。。

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 2000000000
 #define eps 1e-6

 typedef long long ll;
 int dcmp(double x){
    if(x>eps)return ;
    if(x<-eps)return -;
    return ;
 }
 struct point{
    double x,y;
    point(){}
    point(double x,double y):x(x),y(y){}
    point operator - (point a){return point(x-a.x,y-a.y);}
    point operator + (point a){return point(x+a.x,y+a.y);}
    point operator * (double c){return point(x*c,y*c);}
 }po[],fin[];
 struct line{
    point p,v;
    double ang;
    line(){}
    line(point p,point v):p(p),v(v){ang=atan2(v.y,v.x);}
 }li[];
 double cross(point a,point b){
        return a.x*b.y-a.y*b.x;
 }
 double dot(point a,point b){
    return a.x*b.x+a.y*b.y;
 }
 double length(point a){
    return sqrt(dot(a,a));
 }
 bool Onleft(line L,point p){
    return cross(L.v,p-L.p)>;
 }
 bool cmp_l(line a,line b){
    return a.ang<b.ang;
 }
 bool cmp(point a,point b){
    if(dcmp(a.x-b.x))return a.x<b.x;
    return a.y<b.y;
 }
 point GetIntersection(line a,line b){
     point u=a.p-b.p;
     double t=cross(b.v,u)/cross(a.v,b.v);
     return a.p+a.v*t;
 }
 point deq[];
 line q[];
 int HalfPlane(line *L,int n,point *poly){
    sort(L,L+n,cmp_l);

    int l,r;
    q[l=r=]=L[];
    for(int i=;i<n;i++){
       while(l<r && !Onleft(L[i],deq[r-]))r--;
       while(l<r && !Onleft(L[i],deq[l]))l++;
       q[++r]=L[i];
       if(!dcmp(q[r].ang-q[r-].ang)){
          r--;
          if(Onleft(q[r],L[i].p))q[r]=L[i];
       }
       if(l<r)deq[r-]=GetIntersection(q[r],q[r-]);
    }
    while(l<r && !Onleft(q[l],deq[r-]))r--;
    if(r>l+)deq[r]=GetIntersection(q[r],q[l]);
    else return ;

    int m=;
    for(int i=l;i<=r;i++)poly[m++]=deq[i];
    return m;
 }
 int perm[],vis[],h,w;
 double ans;
 void dfs(int now,int n){
    if(now==n){
       li[]=line(point(,),point(,));
       li[]=line(point(w,),point(,));
       li[]=line(point(w,h),point(-,));
       li[]=line(point(,h),point(,-));
       int m=HalfPlane(li,,po);
       double temp=;
       point tp[];
       for(int i=;i<n;i++){
          line now=line(fin[perm[i]],fin[perm[i]]-fin[(perm[i]+)%n]);
          int res=;
          for(int j=;j<m;j++){
             if(!dcmp(cross(now.v,po[(j+)%m]-po[j])))continue;
             tp[res]=GetIntersection(now,line(po[j],po[(j+)%m]-po[j]));
             if(dcmp(dot(tp[res]-po[j],tp[res]-po[(j+)%m]))<=)res++;
          }
          sort(tp,tp+res,cmp);
          if(res>)temp+=length(tp[res-]-tp[]);
          else continue;
          li[i+]=now;
          m=HalfPlane(li,i+,po);
       }
       ans=min(ans,temp);
       return;
    }
    for(int i=;i<n;i++)if(!vis[i]){
       vis[i]=;
       perm[now]=i;
       dfs(now+,n);
       vis[i]=;
    }
 }
 int main(){
     //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
     int t;
     scanf("%d",&t);
     while(t--){
        int vt;
        scanf("%d%d",&w,&h);
        scanf("%d",&vt);
        for(int i=;i<vt;i++)scanf("%lf%lf",&fin[i].x,&fin[i].y);
        ans=INF;
        memset(vis,,sizeof vis);
        dfs(,vt);
        printf("Minimum total length = %.3lf\n",ans);
        if(t)printf("\n");
     }
     return ;
 }

D - Dreisam Equations

就是暴力加表达式计算。（一直SegmentationFault不明所以，发现是代表符号的值取的不好会和数值冲突=_=）

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>
 #include <stack>

 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 200000000
 #define zuokuo 200000000
 #define youkuo 200000001
 #define kong 200000009
 #define jia -200000000
 #define jian -200000001
 #define cheng -200000002

 typedef long long ll;
 char eq[];
 int q[],res;
 int lf;
 stack <int> a;
 stack <int> s;
 int solve(int cur){
    if(cur==res){
       while(!a.empty())a.pop();
       while(!s.empty())s.pop();
       for(int i=;i<res;i++){
          if(q[i]>-&&q[i]<){
             if(s.empty()||s.top()==zuokuo)a.push(q[i]);
             else{
                   int u=a.top();a.pop();
                   if(s.top()==jia)a.push(u+q[i]);
                   if(s.top()==jian)a.push(u-q[i]);
                   if(s.top()==cheng)a.push(u*q[i]);
                   s.pop();
             }
          }
          else if(q[i]==youkuo){
               if(s.empty()||s.top()!=zuokuo)return ;
               s.pop();
               if(s.empty()||s.top()==zuokuo)continue;
               int u=a.top();a.pop();
               int v=a.top();a.pop();
               if(s.top()==jia)a.push(v+u);
               if(s.top()==jian)a.push(v-u);
               if(s.top()==cheng)a.push(v*u);
                     s.pop();
          }
          else s.push(q[i]);
       }
       if(a.top()==lf)return ;
       else return ;
    }
    if(q[cur]!=kong)return solve(cur+);
    for(int i=cheng;i<=jia;i++){
       q[cur]=i;
       if(solve(cur+))return ;
       q[cur]=kong;
    }
    return ;
 }
 int main(){
    int cs=;
    while(gets(eq) && eq[]!=''){
       sscanf(eq,"%d",&lf);
       res=;
       int i;
       for(i=;eq[i]!='=';i++);
       for(i++;eq[i]!='\0';i++){
          for(;eq[i]==' ';i++);
          if(eq[i]=='('){q[res++]=zuokuo;continue;}
          if(eq[i]==')'){q[res-]=youkuo;q[res++]=kong;continue;}
          int num=;
          while(eq[i]>=''&&eq[i]<=''){
             num=num*+eq[i]-'';
             i++;
          }
          q[res++]=num;
          q[res++]=kong;
          i=i-;
       }
       res=res-;

       printf("Equation #%d:\n",cs++);
       if(solve()==)printf("Impossible\n\n");
       else{
          printf("%d=",lf);
          for(int i=;i<res;i++){
             if(q[i]==zuokuo)printf("(");
             else if(q[i]==youkuo)printf(")");
             else if(q[i]==jia)printf("+");
             else if(q[i]==jian)printf("-");
             else if(q[i]==cheng)printf("*");
             else printf("%d",q[i]);
          }
          printf("\n\n");
       }

    }
    return ;
 }