# A. Rudolph and Cut the Rope
只需要按照钉子距离的高度$a_{i}$和绳子的长度$b_{i}$的差值进行排序即可
代码
```
int n;
pii a[N];
bool cmp(pii a,pii b)
{
return a.x-a.y<b.x-b.y;
}
void solve()
{
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i].x>>a[i].y;
sort(a+1,a+1+n,cmp);
reverse(a+1,a+1+n);
int res=0;
for(int i=1;i<=n;i++)
{
if(a[i].x<=a[i].y)
{
break;
}
res++;
}
cout<<res<<endl;
}
```
# B. Rudolph and Tic-Tac-Toe
模拟即可
代码
```
void solve()
{
int n=3;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
cin>>s[i][j];
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(s[i][j]==s[i][j+1]&&s[i][j+1]==s[i][j+2]&&s[i][j]!='.')
{
cout<<s[i][j]<<endl;
return;
}
if(s[i][j]==s[i+1][j]&&s[i+1][j]==s[i+2][j]&&s[i][j]!='.')
{
cout<<s[i][j]<<endl;
return;
}
if(s[i][j]==s[i+1][j+1]&&s[i+1][j+1]==s[i+2][j+2]&&s[i][j]!='.')
{
cout<<s[i][j]<<endl;
return;
}
if(s[i][j]==s[i+1][j-1]&&s[i+1][j-1]==s[i+2][j-2]&&s[i][j]!='.')
{
cout<<s[i][j]<<endl;
return;
}
}
}
cout<<"DRAW"<<endl;
}
```
# C. Rudolf and the Another Competition
统计出每个人的过题的数量$x_{i}$和罚时$t_{i}$然后依次先对比过题数然后再对比罚时即可
```
void solve()
{
cin>>n>>m>>t;
vector<pii>p(n+10);
int res=0;
for(int i=0;i<n;i++)
{
p[i].x=p[i].y=0;
vector<int>a;
int s=0;
for(int j=0;j<m;j++)
{
int x;
cin>>x;
a.push_back(x);
}
sort(a.begin(),a.end());
for(int j=0;j<m;j++)
{
s=s+a[j];
if(s<=t)
{
p[i].x+=1;
p[i].y+=s;
}
}
if(i>0)
{
if((p[0].x<p[i].x)||(p[0].x==p[i].x&&p[i].y<p[0].y))res++;
}
}
cout<<res+1<<endl;
}
```
# D. Rudolph and Christmas Tree
先统计所有三角形的面积，然后减去重叠部分即可，重叠部分可以根据初中学过的相似三角形定理对应边成比例即可
```
void solve()
{
cin>>n>>d>>h;
for(int i=1;i<=n;i++)
cin>>y[i];
sort(y+1,y+1+n);
double s=n*d*h*0.5;
for(int i=2;i<=n;i++)
{
int v=h+y[i-1];
if(v>y[i])
{
double x=v-y[i]*1.0;
double t=x/h*d*x*0.5;
s-=t;
}
}
printf("%f\n",s);
}
```
# E2. Rudolf and Snowflakes (hard version)
根据样例中的图解
节点个数假设为$n$
$n=1+4^1+4^{2}=21$
我们假设他有$k$个子节点，会扩展$x$次那我们可以推出一个方程
$n=1+k+k^{2}+....+k^{x}$
所以只要将$n$分解成多项式即可
我们可以看出这个具有单调性，所以可以利用二分来快速求得答案
先枚举指数$x\in[2,60]$ 然后再二分$k$即可
```
void solve()
{
cin>>n;
for(int i=2;i<=60;i++)
{
int l=1,r=n+1;
auto check=[&](int x){
int s=0,t=1;
for(int j=0;j<=i;j++)
{
s+=t;
if(s>inf)s=inf;
if((__int128_t)t*x>inf)t=inf;
else t*=x;
}
return s;
};
while(l+1<r)
{
int mid=l+r>>1;
if(check(mid)>=n)r=mid;
else l=mid;
}
if(check(r)==n)
{
cout<<"YES"<<endl;
return;
}
}
cout<<"NO"<<endl;
return;
}
int mid=l+r>>1;
if(check(mid)>=n)r=mid;
else l=mid;
}
if(check(r)==n)
{
cout<<"YES"<<endl;
return;
}
}
cout<<"NO"<<endl;
return;
}
```
