【计算几何】【DP】【tarjan】Day 10.6

 1 #include <cstdio>
 2 #include <cmath>
 3 double w,x,r;
 4 int p;
 5 int main()
 6 {
 7     freopen("coin.in","r",stdin);
 8     freopen("coin.out","w",stdout);
 9     scanf("%lf%d",&r,&p);
10     w=r*sqrt(3.0)/3;
11     x=w*2-r;
12     w-=x;
13     for (int i=2;i<=p;i++)
14     {
15         x=pow(w,2.0)/(2*(w+r));
16         w-=2*x;
17     }
18     printf("%.6lf",x);
19 }

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cmath> 
 5 #define ll long long
 6 using namespace std;
 7 ll l,r,f[19][10],w[19],ans=0,ten[19]={1};
 8 ll count(ll x)
 9 {
10     ll s=0;
11     while(x)
12     {
13         s++;
14         w[s]=x%10;
15         x/=10;
16     }
17     return s;
18 }
19 ll check(ll x)//统计小于等于x且和x位数相同的幸运数数量 
20 {
21     memset(f,0,sizeof(f));
22     ll cnt=count(x),s=0,pd=1;
23     if (cnt==0) return 0;
24     for (int i=1;i<w[cnt];i++) f[cnt][i]=1;
25     for (int i=cnt-1;i>=1;i--)//统计小于x且和x位数相同的幸运数数量 
26     {
27         for (int j=0;j<=9;j++)
28         {
29             for (int k=0;k<=9;k++)
30             {
31                 if (abs(j-k)>=2)
32                 {
33                     if (k==w[i+1]&&j<w[i]) f[i][j]+=pd;//从前缀完全相同的状态转移（前提是第i+1位及之前的前缀合法） 
34                     f[i][j]+=f[i+1][k];
35                 }
36             }
37         }
38         if (abs(w[i]-w[i+1])<2) pd=0;//判定第i位及之前的前缀合法 
39     }
40     //统计比x小的幸运数 
41     for (int i=0;i<=9;i++) s+=f[1][i];
42     //统计x本身是不是幸运数 
43     s+=pd;
44     return s;
45 }
46 int main()
47 {
48     freopen("lucky.in","r",stdin);
49     freopen("lucky.out","w",stdout);
50     cin>>l>>r;
51     for (int i=1;i<=17;i++) ten[i]=ten[i-1]*10;
52     for (int i=count(l);i<=count(r)-1;i++) ans+=check(ten[i]-1);
53     ans+=check(r)-check(l-1);
54     cout<<ans<<endl;
55 }