Jiamusi training Day6

T1 still find the law ah, note the number of b, then directly like 10000000009 * 2%

 1 #include <bits/stdc++.h>
 2 #define mo 1000000007
 3 using namespace std;
 4 char s[1000050];
 5 long long b,o;
 6 long long ans;
 7 int main()
 8 {
 9     cin>>s;
10     int length=strlen(s);
11     for(int i=length-1;i>=0;i--)
12     {
13         if(s[i]=='a')
14         o++;
15     }
16     if(o==length)
17     {
18         cout<<"-1"<<endl;
19         return 0;
20     }
21     for(int i=length-1;i>=0;i--)
22     {
23         if(s[i]=='b')
24         {
25             b++;
26             continue;
27         }
28         ans+=b;
29         b=b*2%mo;   
30     }       
31     printf("%lld",ans%mo);
32 } 

View Code

T2 more fun, I hung up the fight

The original idea is to constantly remember binary number, up to five times, five times the conversion, an array of five, then 11 alignments, theory can be had, but I hung up

Positive solution is actually a DP? ? ?

 1 #include <bits/stdc++.h>
 2 #define ll long long 
 3 #define mo 1000000007
 4 using namespace std;
 5 ll b,k,sum,now;
 6 int p,cnt;
 7 ll ans[2000];
 8 char a[2050];
 9 ll dp[2000][2000];
10 void solve(int x)
11 {
12     int point=x;
13     int num=0;
14     while(x)
15     {
16         if(x%2==1)
17         num++;
18         x/=2;
19     }
20     ans[point]=ans[num]+1;
21 }
22 int main()
23 {
24     cin>>a>>k;
25     int length=strlen(a);
26     if(k==1)
27     {
28         cout<<length-1<<endl;   
29         return 0;
30     }
31     if(k==0)
32     {
33         cout<<1<<endl;
34         return 0;
35     }
36     if(k>5)
37     {
38         cout<<0<<endl;
39         return 0;
40     }
41     k--;
42     for(int i=2;i<=1000;i++)
43     solve(i);
44     for(int i=0;i<length;i++)
45     dp[i][0]=1;
46     for(int i=0;i<length;i++)
47     dp[i][i]=1;
48     for(int i=1;i<length;i++)
49     for(int j=1;j<=i;j++)
50     {
51         dp[i][j]=(dp[i-1][j]+dp[i-1][j-1])%mo;
52     }
53     for(int i=0;i<length;i++)
54     {
55         if(a[i]=='0')
56         continue;
57         for(int j=0;j<length-i;j++)
58         {
59             if(ans[j+now]==k)
60             {
61                 sum+=dp[length-i-1][j];
62                 sum=sum%mo;
63             }
64         }
65         now++;
66     }
67     if(ans[now]==k)
68     sum++;
69     cout<<sum%mo<<endl;
70     return 0;
71 }

View Code

T3 practice is part of the sub-tree diameter, 50 theory available, but I hung up ...

 

Solutions are also less likely now, the code first empty 2333