2018-2019 ACM-ICPC, Asia Nanjing Regional Contest
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
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