D Makoto and a Blackboard

Makoto and a Blackboard
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Makoto has a big blackboard with a positive integer nn written on it. He will perform the following action exactly kk times:

Suppose the number currently written on the blackboard is vv. He will randomly pick one of the divisors of vv (possibly 11 and vv) and replace vv with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 5858 as his generator seed, each divisor is guaranteed to be chosen with equal probability.

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

It can be shown that this value can be represented as PQPQ where PP and QQ are coprime integers and Q≢0(mod109+7)Q≢0(mod109+7). Print the value of P⋅Q−1P⋅Q−1 modulo 109+7109+7.

 

 

Input

 

The only line of the input contains two integers nn and kk (1≤n≤10151≤n≤1015, 1≤k≤1041≤k≤104).

 

 

Output

 

Print a single integer — the expected value of the number on the blackboard after kk steps as P⋅Q−1(mod109+7)P⋅Q−1(mod109+7) for PP, QQ defined above.

 

 

Examples

 

 

input
6 1
output
3
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 11, 22, 33 or 66 — each occurring with equal probability. Hence, the answer is 1+2+3+64=31+2+3+64=3.

In the second example, the answer is equal to 1⋅916+2⋅316+3⋅316+6⋅116=1581⋅916+2⋅316+3⋅316+6⋅116=158.

 

 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\n",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

 

posted @ 2019-01-05 16:13  DWVictor  阅读(408)  评论(0)  编辑  收藏  举报