D Makoto and a Blackboard

Makoto has a big blackboard with a positive integer

Suppose the number currently written on the blackboard is

He now wonders what is the expected value of the number written on the blackboard after

It can be shown that this value can be represented as

Input

The only line of the input contains two integers

Output

Print a single integer — the expected value of the number on the blackboard after

Examples

input

6 1

output

input

6 2

output

875000008

input

60 5

output

237178099

Note

In the first example, after one step, the number written on the blackboard is

In the second example, the answer is equal to

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<algorithm>
 7 using namespace std;
 8 const int MAXN=40000000;
 9 const long long mod=1000000007;
10 int k,pri[MAXN];
11 bool exist[MAXN];
12 long long n,inv[200],f[10050][200];
13 long long pOw(long long a,long long m)
14 {
15     long long pro;
16     for(pro=1LL;m;m>>=1,a=a*a%mod)
17         if(m&1)
18             pro=pro*a%mod;
19     return pro;
20 }
21 void pre_calc()
22 {
23     memset(exist,true,sizeof(exist));
24     pri[0]=0;
25     for(int i=2;i<MAXN;i++)
26     {
27         if(exist[i]) pri[++pri[0]]=i;
28         for(int j=1;j<=pri[0]&&(long long)i*pri[j]<MAXN;j++)
29         {
30             exist[i*pri[j]]=false;
31             if(i%pri[j]==0)
32                 break;
33         }
34     }
35     inv[0]=1;
36     for(int i=1;i<200;i++)
37         inv[i]=pOw(i,mod-2);
38     return;
39 }
40 long long calc(long long p,int num)
41 {
42     f[0][0]=1LL;
43     for(int i=1;i<=num;i++)
44         f[0][i]=f[0][i-1]*p%mod;
45     for(int t=1;t<=k;t++)
46     {
47         f[t][0]=f[t-1][0];
48         for(int i=1;i<=num;i++)
49             f[t][i]=(f[t][i-1]+f[t-1][i])%mod;
50         for(int i=1;i<=num;i++)
51             f[t][i]=f[t][i]*inv[i+1]%mod;
52     }
53     return f[k][num];
54 }
55 int main()
56 {
57     int num;
58     pre_calc();
59     cin>>n>>k;
60     long long ans=1LL;
61     for(int i=1;i<=pri[0]&&(long long)pri[i]*pri[i]<=n;i++) if(n%pri[i]==0)
62     {
63         for(num=0;n%pri[i]==0;n/=pri[i],num++);
64         ans=ans*calc(pri[i],num)%mod;
65     }
66     if(n!=1) ans=ans*calc(n,1)%mod;
67     cout<<ans;
68     fclose(stdin);
69     fclose(stdout);
70     return 0;
71 }

View Code

学习一下

Dinic+当前弧优化

网络流的dinic算法详解以及当前弧优化备注：点开

 1 /*
 2     最大流 Dinic 
 3   
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstdlib>
 8 #include<string>
 9 #include<queue>
10 #include<cstring>
11 #define min(x,y)    ((x<y)?(x):(y))
12 #define rev(i)(i&1?(i+1):(i-1))
13 using namespace std;
14 typedef long long ll;
15 int dis[10005]; //分层图，距源点距离
16 int cur[10005]; //当前弧优化 
17 int n,m,ans,st,ed;
18 struct node{
19     int x,y,len,nxt;
20     node(){}
21     node(int nx,int ny,int nlen,int nnxt){
22         x=nx;y=ny;len=nlen;nxt=nnxt;        
23     }
24 } E[200010];
25 int head[10001],cnt;
26 int bfs(){
27     for (int i=1;i<=n;i++)  dis[i]=-1;
28     queue<int> Q;
29     dis[st]=0;Q.push(st);
30     while (!Q.empty()){
31         int j=Q.front();
32         Q.pop();
33         for (int i=head[j];i;i=E[i].nxt)
34             if (dis[E[i].y]<0&&E[i].len>0){
35                 dis[E[i].y]=dis[j]+1;
36                 Q.push(E[i].y);
37             }
38     }
39     if (dis[ed]>0)  return 1;
40     else return 0;
41 }
42 int find(int x,int low){
43     int res=0;
44     if (x==ed)  return low;
45     for (int i=cur[x];i;i=E[i].nxt){
46     cur[x]=i;
47     if (E[i].len>0&&dis[E[i].y]==dis[x]+1&&(res=find(E[i].y,min(low,E[i].len))))    
48         {
49             E[i].len-=res;
50             E[i^1].len+=res;
51             return res;
52         }
53     }
54     return 0;
55 }
56 inline void link(int x,int y,int z){
57     E[++cnt]=node(x,y,z,head[x]);
58     head[x]=cnt;
59     E[++cnt]=node(y,x,0,head[y]);
60     head[y]=cnt;
61 }
62 int main(){
63     scanf("%d%d%d%d",&n,&m,&st,&ed);
64     cnt=1;  
65     for (int i=1;i<=m;i++)
66     {
67         int a,b,c;
68         scanf("%d%d%d",&a,&b,&c);
69         link(a,b,c);
70     }
71     ans=0;int tans=0;
72     while(bfs()){
73         for (int i=1;i<=n;i++)  cur[i]=head[i];
74         while (tans=find(st,2e6))   ans+=tans;
75     }
76 
77     printf("%d
",ans);
78     return 0;
79 }

View Code

网络流

https://www.cnblogs.com/SYCstudio/p/7260613.html

经典的最大流题POJ1273

http://poj.org/problem?id=1273