https://codeforces.com/gym/101981

Problem A. Adrien and Austin

贪心，注意细节

f[x]=1：先手必赢。

f[x]: 分成两部分(或一部分)，长度分别为a和b，只要存在f[a] xor f[b]=0，则f[x]=1。

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define minv 1e-6

 #define inf 1e9

 #define pi 3.1415926536

 #define nl 2.7182818284

 const ll mod=1e9+;//

 const int maxn=1e5+;

 int main()

 {

     int n,k;

     scanf("%d%d",&n,&k);

     if (k==)

     {

         if (n & )

             printf("Adrien");

         else

             printf("Austin");

     }

     else if (n==)

         printf("Austin");

     else

         printf("Adrien");

     return ;

 }

Problem D. Country Meow

三分套三分套三分

某种三分的写法是不行的！(x x+minv 精度不够)

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alt="" />

拓展：

in k-dimension Cartesian coordinate, find a point that sum of Euclidean distance from some points to it is smallest

k次三分

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define minv 1e-8

 #define inf 1e9

 #define pi 3.1415926536

 #define nl 2.7182818284

 const ll mod=1e9+;//

 const int maxn=1e2+;

 double dx[maxn],dy[maxn],dz[maxn],value=1e-;

 int n;

 double Euc(int i,double xx,double yy,double zz)

 {

     return sqrt((dx[i]-xx)*(dx[i]-xx) + (dy[i]-yy)*(dy[i]-yy) + (dz[i]-zz)*(dz[i]-zz));

 }

 double dist(double xx,double yy,double zz)

 {

     double re=;

     int i;

     for (i=;i<=n;i++)

         re=max(re,Euc(i,xx,yy,zz));

     return re;

 }

 double work3(double x,double y)

 {

     double zl=-,zr=,z1,z2,d1,d2;

     while (zr-zl>value)

     {

         z1=(zl+zr)/;

         z2=(z1+zr)/;

         d1=dist(x,y,z1);

         d2=dist(x,y,z2);

         if (d1>d2)

             zl=z1;

         else

             zr=z2;

     }

     return min(d1,d2);

 }

 double work2(double x)

 {

     double yl=-,yr=,y1,y2,d1,d2;

     while (yr-yl>value)

     {

         y1=(yl+yr)/;

         y2=(y1+yr)/;

         d1=work3(x,y1);

         d2=work3(x,y2);

         if (d1>d2)

             yl=y1;

         else

             yr=y2;

     }

     return min(d1,d2);

 }

 double work1()

 {

     double xl=-,xr=,x1,x2,d1,d2;

     while (xr-xl>value)

     {

         x1=(xl+xr)/;

         x2=(x1+xr)/;

         d1=work2(x1);

         d2=work2(x2);

         if (d1>d2)

             xl=x1;

         else

             xr=x2;

     }

     return min(d1,d2);

 }

 int main()

 {

     int i;

     scanf("%d",&n);

     for (i=;i<=n;i++)

         cin>>dx[i]>>dy[i]>>dz[i];

     printf("%.10f",work1());

     return ;

 }

Problem E. Eva and Euro coins

看错题意了，难受！

Problem G. Pyramid
急需队友协助。。。

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define minv 1e-6

 #define inf 1e9

 #define pi 3.1415926536

 #define nl 2.7182818284

 const ll mod=1e9+;//

 const int maxn=1e5+;

 ll pow(ll a,ll b)

 {

     ll y=;

     while (b)

     {

         if (b & )

             y=y*a%mod;

         a=a*a%mod;

         b>>=;

     }

     return y;

 }

 int main()

 {

     int t;

     ll sum,v,vv,n,ni2=pow(2ll,mod-),ni3=pow(3ll,mod-),ni4=pow(4ll,mod-),ni6=pow(6ll,mod-);

     scanf("%d",&t);

     while (t--)

     {

         scanf("%lld",&n);

         sum=;

         sum=(sum + n*(n+)%mod*(n+)%mod*ni6 )%mod;

         v=;

         v=(v + (n-)*n%mod*(n+)%mod*ni6 )%mod;

         if (n & )

             v=(v + (n*n-)%mod*ni4 )%mod;

         else

             v=(v + (n*n)%mod*ni4 )%mod;

         sum=(sum + v*ni2 )%mod;

         v=;

         v=(v + (n-)*(n-)%mod*n%mod*(n+)%mod*ni4 )%mod;

         vv=;

         vv=(vv + (n-)*(n-)%mod*n%mod*ni6 )%mod;

         if (n & )

             vv=(vv + (n-)*(n-)%mod*ni4 )%mod;

         else

             vv=(vv + ((n-)*(n-)-)%mod*ni4 )%mod;

         v=(v + vv*)%mod;

         sum=(sum + v*ni6 )%mod;

         printf("%lld\n",(sum+mod)%mod);

     }

     return ;

 }

Problem I. Magic Potion

网络最大流。

差点以为是上下界网络流，看到这么多人对，感觉特别诧异。。。

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define minv 1e-6

 #define inf 1e9

 #define pi 3.1415926536

 #define nl 2.7182818284

 const ll mod=1e9+;//

 const int maxn=1e3+;

 struct node

 {

     int d,len;

     node *next,*opp;

 }*e[maxn],*pre[maxn];

 queue<int> st;

 int s,t,sum=,add[maxn];

 bool vis[maxn];

 void add_edge(int x,int y,int len)

 {

     node *p1=(node*) malloc (sizeof(node));

     node *p2=(node*) malloc (sizeof(node));

     p1->d=y;

     p1->len=len;

     p1->next=e[x];

     p1->opp=p2;

     e[x]=p1;

     p2->d=x;

     p2->len=;

     p2->next=e[y];

     p2->opp=p1;

     e[y]=p2;

 }

 void bfs()

 {

     int d,dd,v;

     node *p;

     while ()

     {

         memset(vis,,sizeof(vis));

         memset(add,,sizeof(add));

         add[s]=inf;

         vis[s]=;

         st.push(s);

         while (!st.empty())

         {

             d=st.front();

             st.pop();

             p=e[d];

             while (p)

             {

                 dd=p->d;

                 v=min(add[d],p->len);

                 if (add[dd]<v)

                 {

                     add[dd]=v;

                     pre[dd]=p->opp;

                     if (!vis[dd])

                     {

                         vis[dd]=;

                         st.push(dd);

                     }

                 }

                 p=p->next;

             }

             vis[d]=;

         }

         if (add[t]==)

             return;

         sum+=add[t];

         d=t;

         while (d!=s)

         {

             pre[d]->len+=add[t];

             pre[d]->opp->len-=add[t];

             d=pre[d]->d;

         }

     }

 }

 int main()

 {

     int n,m,k,d,g,i;

     scanf("%d%d%d",&n,&m,&k);

     s=n+m+,t=n+m+;

     add_edge(s,,k);

     for (i=;i<=n;i++)

     {

         add_edge(s,i,);

         add_edge(,i,);

     }

     for (i=n+;i<=n+m;i++)

         add_edge(i,t,);

     for (i=;i<=n;i++)

     {

         scanf("%d",&g);

         while (g--)

         {

             scanf("%d",&d);

             add_edge(i,n+d,);

         }

     }

     bfs();

     printf("%d",sum);

     return ;

 }

Problem J. Prime Game

预处理一个数的所有质因数。

同一个质因数，分段处理。

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define minv 1e-6

 #define inf 1e9

 #define pi 3.1415926536

 #define nl 2.7182818284

 const ll mod=1e9+;//

 const int maxn=1e6+;

 int zhi[maxn],zys[maxn][],s[maxn],a[maxn];

 bool vis[maxn];

 int main()

 {

     int n,g=,value=1e6,i,j,k,l;

     ll sum=;

     for (i=;i<=value;i++)

     {

         if (!vis[i])

         {

             zhi[++g]=i;

             zys[i][]=;

             zys[i][]=g;

         }

         for (j=;j<=g;j++)

         {

             k=i*zhi[j];

             if (k>value)

                 break;

             if (i%zhi[j]==)

             {

                 zys[k][]=zys[i][];

                 for (l=;l<=zys[i][];l++)

                     zys[k][l]=zys[i][l];

             }

             else

             {

                 zys[k][]=zys[i][]+;

                 zys[k][]=j;

                 for (l=;l<=zys[i][];l++)

                     zys[k][l+]=zys[i][l];

             }

             vis[k]=;

             if (i%zhi[j]==)

                 break;

         }

     }

     scanf("%d",&n);

     for (i=;i<=n;i++)

         scanf("%d",&a[i]);

     for (i=;i<=n;i++)

     {

         k=a[i];

         for (j=;j<=zys[k][];j++)

         {

             sum+=1ll*(i-s[zys[k][j]])*(n-i+);

             s[zys[k][j]]=i;

         }

     }

     printf("%lld",sum);

     return ;

 }

Problem K. Kangaroo Puzzle

一旦两个数在一起，那么两个数之后永远都在一起

神仙方法！（看n,m数据那么小，操作次数那么多……）

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <ctime>

 #include <cstring>

 #include <string>

 #include <map>

 #include <set>

 #include <list>

 #include <queue>

 #include <stack>

 #include <vector>

 #include <bitset>

 #include <algorithm>

 #include <iostream>

 using namespace std;

 #define ll long long

 const int maxn=1e5+;

 char str[]="LRUD";

 int main()

 {

     int g=;//

     srand(time(NULL));

     for (int i=;i<=g;i++)

         printf("%c",str[rand()%]);

     return ;

 }

Problem M. Mediocre String Problem

以待后续埋坑

网络赛也有一题Manacher的