Divide by Zero 2017 D&E&F

Posted 2020-09-22 弥生三月

唔，D：概率最多是 $ \frac{1}{2} $ ，在n==1000的时候这个阈值大概是7284，O(7284n)即可

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <string>
 6 #include <cstring>
 7 #include <cmath>
 8 #include <map>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <queue>
13 #include <time.h>
14 #define eps 1e-7
15 #define INF 0x3f3f3f3f
16 #define MOD 1000000007
17 #define rep0(j,n) for(int j=0;j<n;++j)
18 #define rep1(j,n) for(int j=1;j<=n;++j)
19 #define pb push_back
20 #define set0(n) memset(n,0,sizeof(n))
21 #define ll long long
22 #define ull unsigned long long
23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
24 #define max(a,b) (a>b?a:b)
25 #define min(a,b) (a<b?a:b)
26 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
27 #define TO(j) printf(#j": %d\n",j);
28 //#define OJ
29 using namespace std;
30 const int MAXINT = 100010;
31 const int MAXNODE = 100010;
32 const int MAXEDGE = 2*MAXNODE;
33 char BUF,*buf;
34 int read(){
35     char c=getchar();int f=1,x=0;
36     while(!isdigit(c)){if(c==‘-‘) f=-1;c=getchar();}
37     while(isdigit(c)){x=x*10+c-‘0‘;c=getchar();}
38     return f*x;
39 }
40 char get_ch(){
41     char c=getchar();
42     while(!isalpha(c)) c=getchar();
43     return c;
44 }
45 //------------------- Head Files ----------------------//
46 int n,k;
47 double dp[7300][1001],prop[20000];
48 
49 void get_input();
50 void work();
51 int main() {
52     get_input();
53     work();
54     return 0;
55 }
56 void work(){
57     dp[0][0] = 1;
58     for(int i=1;i<7300;i++){
59         for(int j=0;j<=n;j++){
60             dp[i][j]=dp[i-1][j]*j/double(n);
61             if(j!=0) dp[i][j]+=dp[i-1][j-1]*(n-j+1)/double(n);
62         }
63     }
64     rep0(i,7300) prop[i]=dp[i][n];
65     while(k--){
66         int p = read();
67         printf("%d\n",lower_bound(prop,prop+7300,(p-eps)/2000.0)-prop);
68     }
69 }
70 void get_input(){
71     n=read();k=read();
72 }

D：少女祈祷中

E：考虑一下最原始的nim游戏里每堆石子数量的意义，它的意思是这堆石子可以在1,2,3..n次内取完，并且这个信息支持减法，然后这题里的这个规则好像也是这个意思，事实上每一个有这种限制的石碓可以映射到一个原来的石碓上，即 $ sg(n) = t (\sum_{i=1}^{t} \leqslant n < \sum_{i=1}^{t+1}) $ ，然后直接异或起来即可

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <string>
 6 #include <cstring>
 7 #include <cmath>
 8 #include <map>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <queue>
13 #include <time.h>
14 #define eps 1e-7
15 #define INF 0x3f3f3f3f
16 #define MOD 1000000007
17 #define rep0(j,n) for(int j=0;j<n;++j)
18 #define rep1(j,n) for(int j=1;j<=n;++j)
19 #define pb push_back
20 #define set0(n) memset(n,0,sizeof(n))
21 #define ll long long
22 #define ull unsigned long long
23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
24 #define max(a,b) (a>b?a:b)
25 #define min(a,b) (a<b?a:b)
26 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
27 #define TO(j) printf(#j": %d\n",j);
28 //#define OJ
29 using namespace std;
30 const int MAXINT = 100010;
31 const int MAXNODE = 100010;
32 const int MAXEDGE = 2*MAXNODE;
33 char BUF,*buf;
34 int read(){
35     char c=getchar();int f=1,x=0;
36     while(!isdigit(c)){if(c==‘-‘) f=-1;c=getchar();}
37     while(isdigit(c)){x=x*10+c-‘0‘;c=getchar();}
38     return f*x;
39 }
40 char get_ch(){
41     char c=getchar();
42     while(!isalpha(c)) c=getchar();
43     return c;
44 }
45 //------------------- Head Files ----------------------//
46 
47 int sg[500],v,n;
48 void get_input();
49 void work();
50 int main() {
51     get_input();
52     work();
53     return 0;
54 }
55 void work(){
56     if(v) printf("NO\n"); else printf("YES\n");
57 }
58 void get_input(){
59     for(int i=1;i<13;i++){
60         for(int j=i*(i+1)/2;j<(i+2)*(i+1)/2;j++){
61             sg[j]=i;
62         }
63     }
64     n=read();
65     rep0(i,n){
66         int t=read();
67         v^=sg[t];
68     }
69 }

E：少女祈祷中

F：唔，枚举架子的数量，然后对于某一个确定的架子数量n，分类大力讨论，插板法计数即可，总数直接插板，合法的数量就是先从w里减掉堆数*h再插板

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <string>
 6 #include <cstring>
 7 #include <cmath>
 8 #include <map>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <queue>
13 #include <time.h>
14 #define eps 1e-7
15 #define INF 0x3f3f3f3f
16 #define MOD 1000000007
17 #define rep0(j,n) for(int j=0;j<n;++j)
18 #define rep1(j,n) for(int j=1;j<=n;++j)
19 #define pb push_back
20 #define set0(n) memset(n,0,sizeof(n))
21 #define ll long long
22 #define ull unsigned long long
23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
24 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
25 #define TO(j) printf(#j": %d\n",j);
26 //#define OJ
27 using namespace std;
28 const int MAXINT = 100010;
29 const int MAXNODE = 100010;
30 const int MAXEDGE = 2*MAXNODE;
31 char BUF,*buf;
32 int read(){
33     char c=getchar();int f=1,x=0;
34     while(!isdigit(c)){if(c==‘-‘) f=-1;c=getchar();}
35     while(isdigit(c)){x=x*10+c-‘0‘;c=getchar();}
36     return f*x;
37 }
38 char get_ch(){
39     char c=getchar();
40     while(!isalpha(c)) c=getchar();
41     return c;
42 }
43 //------------------- Head Files ----------------------//
44 
45 ll cnt_t,cnt_g;
46 ll fact[100010],df[100010];
47 ll f,w,h;
48 ll fp(ll b,ll u){
49     ll ans = 1;
50     for(;u;b=b*b%MOD,u>>=1) if(u&1) ans=ans*b%MOD;
51     return ans;
52 }
53 ll C(ll n,ll m){
54     if(m>n) return 0;
55     return (fact[n]*df[m]%MOD)*df[n-m]%MOD;
56 }
57 void add(ll &a,ll b){
58     a=(a+b)%MOD;
59 }
60 void get_input();
61 void work();
62 int main() {
63     get_input();
64     work();
65     return 0;
66 }
67 void work(){
68     fact[0]=1;
69     rep1(i,100000) fact[i]=fact[i-1]*i%MOD;
70     rep0(i,100001) df[i] = fp(fact[i],MOD-2);
71     //printf("%lld\n",df[0]);
72     for(int i=1;1;i++){
73         if(i>f+w||i>min(f,w)*2+1) break;
74 
75         if(i==1){
76             if(f==0||w==0) cnt_t++;
77             if(w==0||(f==0&&w>h)) cnt_g++;
78         }else{
79             if(i&1){
80                 add(cnt_t,C(f-1,i/2)*C(w-1,i/2-1));
81                 add(cnt_t,C(f-1,i/2-1)*C(w-1,i/2));
82                 add(cnt_g,C(f-1,i/2)*C(w-1-i/2*h,i/2-1));
83                 add(cnt_g,C(f-1,i/2-1)*C(w-1-(i/2+1)*h,i/2));
84             }else{
85                 add(cnt_t,(C(f-1,i/2-1)*C(w-1,i/2-1)%MOD)*2);
86                 add(cnt_g,(C(f-1,i/2-1)*C(w-1-(i/2)*h,i/2-1)%MOD)*2);
87             }
88         }
89         //printf("%d %lld\n",i,cnt_t);
90     }
91     //printf("%lld\n",cnt_t);
92     printf("%lld\n",cnt_g*fp(cnt_t,MOD-2)%MOD);
93 }
94 void get_input(){
95     f=read();w=read();h=read();
96 }

F：少女祈祷中