Codeforces Gym100495 B、D、E、F、K

http://codeforces.com/gym/100495

K题 草地的面积减去相交的面积，计算几何，垃圾题，避免不必要的计算损失精度（能约分的约分）

卡了老子一个星期了 再加前几天的一道题 这一星期真的是难受 什么都没干

AC代码

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 using namespace std;
 const int maxn = 1e3+;
 const int maxm = 1e4+;
 const int inf = 0x3f3f3f3f;
 const int mod = ;
 const double epx = 1e-;
 typedef long long ll;
 const ll INF = 1e18;
 const double pi = acos(-1.0);
 int main()
 {
     int t,kase=;
     double x1,y1,r1,x2,y2,r2;
     cin>>t;
     while(t--)
     {
         scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&r1,&x2,&y2,&r2);
         double len=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
         double ans;
         if(len>=r1+r2)
         {
             ans=r1*r1*pi;
         }
         else if(fabs(r1-r2)>=len)
         {
             if(r1>r2)
                 ans=pi*r1*r1-pi*r2*r2;
             else
                 ans=;
         }
         else
         {
             double a1=acos((len*len+r1*r1-r2*r2)/(*len*r1));
             double a2=acos((len*len+r2*r2-r1*r1)/(*len*r2));
             double area1=a1*r1*r1;
             double area2=a2*r2*r2;
             double area3=len*r1*sin(a1);
             ans=pi*r1*r1-(area1+area2-area3);
         }
         printf("Case #%d: %.8lf\n",kase++,ans);
     }
 }
 close

给出被卡一万年的K的代码

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 using namespace std;
 const int maxn = 1e3+;
 const int maxm = 1e4+;
 const int inf = 0x3f3f3f3f;
 const int mod = ;
 const double epx = 1e-;
 typedef long long ll;
 const ll INF = 1e18;
 const double pi = acos(-1.0);
 int main()
 {
     int t,kase=;
     double x1,y1,r1,x2,y2,r2;
     cin>>t;
     while(t--)
     {
         scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&r1,&x2,&y2,&r2);
         double len=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
         double ans;
         if(len>=r1+r2)
         {
             ans=r1*r1*pi;
         }
         else if(fabs(r1-r2)>=len)
         {
             if(r1>r2)
                 ans=pi*r1*r1-pi*r2*r2;
             else
                 ans=;
         }
         else
         {
             double a1=2.0*acos((len*len+r1*r1-r2*r2)/(*len*r1));
             double a2=2.0*acos((len*len+r2*r2-r1*r1)/(*len*r2));
             double area1=0.5*a1*r1*r1;
             double area2=0.5*a2*r2*r2;
             double area3=len*r1*sin(a1);
             ans=pi*r1*r1-(area1+area2-area3);
         }
         printf("Case #%d: %.11lf\n",kase++,ans);
     }
 }

B题  把每个字符串开头和结尾中间的字符排序，用map标记一下 然后在判断是否出现过就好了

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 using namespace std;
 const int maxn = 1e3+;
 const int maxm = 1e4+;
 const int inf = 0x3f3f3f3f;
 const int mod = ;
 const double epx = 1e-;
 typedef long long ll;
 const ll INF = 1e18;
 const double pi = acos(-1.0);
 string s;
 int ans[maxn];
 int main()
 {
     int n,m,t,kase=;
     cin>>t;
     while(t--)
     {
         map<string,int> mp;
         cin>>n>>m;
         for(int i=;i<n;i++)
         {
             cin>>s;
             int len=s.length();
             if(len>)
                 sort(s.begin()+,s.end()-);
             //cout<<len<<" "<<s<<endl;
             mp[s]++;
         }
         for(int i=;i<m;i++)
         {
             cin>>s;
             int len=s.length();
             if(len>)
                 sort(s.begin()+,s.end()-);
             //cout<<len<<" "<<s<<endl;
             if(mp[s]>)
                 ans[i]=;
             else
                 ans[i]=;
         }
         printf("Case #%d: ",kase++);
         for(int i=;i<m;i++)
             cout<<ans[i];
         cout<<endl;
     }
 }

E题  可以推出要求子串满足min+a>=max 的最大长度  尺取法 右端点从左端点开始推进  直到右端点不符合条件

　　　再从左端点开始到右端点找最值 并且不断推进左端点 直到满足条件 再向右推进

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 using namespace std;
 const int maxn = 1e5+;
 const int maxm = 1e4+;
 const int inf = 0x3f3f3f3f;
 const int mod = ;
 const double epx = 1e-;
 typedef long long ll;
 const ll INF = 1e18;
 int s[maxn];
 int main()
 {
     int t,a,n,kase=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d %d",&n,&a);
         memset(s,,sizeof(s));
         for(int i=;i<=n;i++)
             scanf("%d",&s[i]);
         int len=;
         int l=,maxx,minn;
         maxx=minn=s[l];
         for(int r=;r<=n;)
         {
             if(minn+a>=maxx)
             {
                 len=max(len,r-l+);
                 r++;
                 if(r>n)
                     break;
                 maxx=max(maxx,s[r]);
                 minn=min(minn,s[r]);
             }
             else
             {
                 if(s[r]==maxx)
                 {
                     while(minn+a<maxx)
                     {
                         int tempmin=s[r];
                         for(int i=l+;i<=r;i++)
                             tempmin=min(tempmin,s[i]);
                         minn=tempmin;
                         l++;
                     }
                 }
                 else if(s[r]==minn)
                 {
                     while(minn+a<maxx)
                     {
                         int tempmaxx=s[r];
                         for(int i=l+;i<=r;i++)
                             tempmaxx=max(tempmaxx,s[i]);
                         maxx=tempmaxx;
                         l++;
                     }
                 }
             }
             //cout<<"l="<<l<<" r="<<r<<" min="<<minn<<" max="<<maxx<<endl;
         }
         printf("Case #%d: %d\n",kase++,len);
     }
 }
 close

D题 水题 边乘边取模

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 using namespace std;
 const int maxn = 1e6+;
 const int maxm = 1e4+;
 const int inf = 0x3f3f3f3f;
 const int mod = ;
 const double epx = 1e-;
 typedef long long ll;
 const ll INF = 1e18;
 ll n,m;
 char a[maxn];
 int sum[maxn];
 int sushu(int x)
 {
    for(int i=;i<n;i++)
         if(x%i==)
             return ;
    return ;
 }
 int poww(int x,int y,int z)
 {
     int ans=;
     for(int i=;i<=y;i++)
         ans=(ans*x)%n;
     return ans;
 }
 int main()
 {
     int t;
     cin>>t;
     int kase=;
     while(t--)
     {
         cin>>n;
         if(n==)
             printf("Case #%d: %d\n",kase,n);
         else if(sushu(n))
         {
             printf("Case #%d: %d\n",kase,poww(,n-,n)%n);
         }
         else
             printf("Case #%d: %d\n",kase,poww(n-,,n)%n);
         kase++;
     }
 }
 close

F题  一遍dfs回溯，记录最大深度就好了

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 using namespace std;
 const int maxn = 1e3+;
 const int maxm = 1e4+;
 const int inf = 0x3f3f3f3f;
 const int mod = ;
 const double epx = 1e-;
 typedef long long ll;
 const ll INF = 1e18;
 char a[maxn][maxn];
 int visit[maxn][maxn];
 int n,m,k;
 int sum,flag;
 int dire[][]= {{,},{-,},{,-},{,}};
 void dfs(int x,int y)
 {
     visit[x][y]=;
     for(int i=; i<; i++)
     {
         int x1=(x+dire[i][]+n)%n;
         int y1=(y+dire[i][]+m)%m;
         //printf("x=%d y=%d x1=%d y1=%d sum=%d c=%c\n",x,y,x1,y1,sum,a[x1][y1]);
         if(visit[x1][y1]==&&a[x1][y1]!='#'&&a[x1][y1]!='x')
         {
             if(a[x1][y1]=='o')
             {
                 dfs(x1,y1);
                 visit[x1][y1]=;
             }
             else if(a[x1][y1]=='.')
             {
                 sum++;
                 if(sum>=k)
                     flag=;
                 dfs(x1,y1);
                 visit[x1][y1]=;
                 sum--;
             }
         }
     }
 }
 int main()
 {
     int t,kase=;
     cin>>t;
     while(t--)
     {
         cin>>n>>m>>k;
         for(int i=; i<n; i++)
             cin>>a[i];
         int x,y;
         for(int i=; i<n; i++)
             for(int j=; j<m; j++)
                 if(a[i][j]=='x')
                     x=i,y=j;
 //        cout<<x<<" "<<y<<endl;
 //        for(int i=0;i<4;i++)
 //            cout<<dire[i][0]<<" "<<dire[i][1]<<endl;
         memset(visit,,sizeof(visit));
         flag=;sum=;dfs(x,y);
 //        for(int i=0;i<n;i++)
 //        {
 //            for(int j=0;j<m;j++)
 //            {
 //                cout<<visit[i][j]<<" ";
 //            }
 //            cout<<endl;
 //        }
         if(flag==)
             printf("Case #%d: Fits perfectly!\n",kase++);
         else
             printf("Case #%d: Oh no, snake's too fat!\n",kase++);
     }
 }