**Kyoya and Colored Balls

 Kyoya and Colored Balls

Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color ibefore drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen.

Input

The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.

Then, k lines will follow. The i-th line will contain ci, the number of balls of the i-th color (1 ≤ ci ≤ 1000).

The total number of balls doesn't exceed 1000.

Output

A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007.

Sample test(s)

input

3
2
2
1

output

3

input

4
1
2
3
4

output

1680

Note

In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:

1 2 1 2 3
1 1 2 2 3
2 1 1 2 3

-------------------------------editorial---------------------------------------

this  problem can easily implement if we do some special considerations

initially ans=1; 

 lets decide answer for   using one number at a time 

  let us take  biggest element initially   one of the biggest element must be at the last of the sequence ,soo after fixing one of the bigest number at the last position , now we have pos-1 positions remain and occurance of bigest nu-1 no of biggest number remain so we can arrange those number at any remaining positions  soo  

 ans*=(pos-1) C(no of remaining biggest numger)

after fixing all biggest number number of remaining positions are total positions - number of occurance of biggest number 

now we set answer for second biggest number in the same manar and multiply with the privious ansewer 

------------------------------------------code--------------------------------

#include<bits/stdc++.h>

using namespace std;

typedef long long int lli;

typedef long int li;

typedef unsigned long long int ulli;

#define ff first

#define ss second

#define pb push_back

#define mp make_pair

#define mod 1000000007

#define test() int t, cin>>t,while(t--)

lli combination()

  {

  for(int i = 0; i < 1000; i++) {

        comb[i][0] = 1;

        for(int j = 1; j <= i; j++)

            comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % mod;

    }

  }

  

using namespace std;

int main()

{

int n;

combination();

cin>>n;

 

  vector<int> v;

int tot=0;

for(int i=0;i<n;i++)

{

int d;

cin>>d;

v.push_back(d);

tot+=v[i];

}

 

lli ans=1;

 

 

for(int i=n-1;i>=0;i--)

 {

 

      int count=v[i];

      ans=(ans*comb[tot-1][count-1])%mod;

       tot-=count;

 }

 

 

 cout<<ans<<endl;

 

}

**Magic Five

 Magic Five

There is a long plate s containing n digits. Iahub wants to delete some digits (possibly none, but he is not allowed to delete all the digits) to form his "magic number" on the plate, a number that is divisible by 5. Note that, the resulting number may contain leading zeros.

Now Iahub wants to count the number of ways he can obtain magic number, modulo 1000000007 (109 + 7). Two ways are different, if the set of deleted positions in s differs.

Look at the input part of the statement, s is given in a special form.

Input

In the first line you're given a string a (1 ≤ |a| ≤ 105), containing digits only. In the second line you're given an integer k (1 ≤ k ≤ 109). The plate s is formed by concatenating k copies of a together. That is n = |a|·k.

Output

Print a single integer — the required number of ways modulo 1000000007 (109 + 7).

Sample test(s)

input

1256
1

output

4

input

13990
2

output

528

input

555
2

output

63

Note

In the first case, there are four possible ways to make a number that is divisible by 5: 5, 15, 25 and 125.

In the second case, remember to concatenate the copies of a. The actual plate is 1399013990.

In the third case, except deleting all digits, any choice will do. Therefore there are 26 - 1 = 63 possible ways to delete digits.

--------------------------------editorial----------------------------------------

Property: A number is divisible by 5 if and only if its last digit is either 0 or 5.

A first solution: Suppose you’re given a plate S, not so big, so we can iterate all its elements. Can we get the answer? I build a new array sol[]. In explanation, both S and sol will be 1-based. Denote N = size of S. Also, denote sol[i] = the number of ways to delete digits from plate S such as we obtain a magic number which has the last digit on position i. The answer is sol[1] + sol[2] + … + sol[N]. Let’s focus now on calculating sol[i]. If S[i] (digit of the plate corresponding to ith position) is different than 0 or 5, then sol[i] is 0 (see “property”). Otherwise we have to ask ourself: in how many ways I can delete digits in “left” and in “right” of position i. In the “right”, we have only one way: delete all digits (if one digit from right still stands, then the number isn’t ending at position i). Now in the “left”: there are digits on positions 1, 2, …, i – 1. We can either delete a digit or keep it – anyhow he’d do we still get a magic number. So on position 1 I have 2 ways (delete or keep it), on position 2 I have also 2 ways, …, on position i – 1 I have also 2 ways. Next, we apply what mathematics call “rule of product” and we get 2 * 2 * 2 … * 2 (i – 1 times) = 2 ^ (i – 1). Applying “rule of product” on both “left” and “right” I get 2 ^ (i – 1) * 1 = 2 ^ (i – 1). To sum it up: If S[i] is 0 or 5 we add to the answer 2 ^ (i – 1). Otherwise, we add nothing. The only problem remained for this simple version is how we calculate A ^ B modulo one number. This is a well known problem as well, called “Exponentiation by squaring”.

Coming back to our problem: So what’s different in our problem? It’s the fact that we can’t iterate all elements of plate. However, we can use “concatenation” property. We know that if an element is a position i in the first copy, it will also be on positions i + n, i + 2 * n, i + 3 * n, …, i + (k – 1) * n (we don’t call here about trivial case when k = 1). What if iterate only one copy and calculate for all K copies. If in the first copy, at the position i is either 0 or 5, we have to calculate the sum 2 ^ i + 2 ^ (i + n) + 2 ^ (i + 2 * n) + … + 2 ^ (i + (k – 1) * n). By now on, in calculus I'll denote i as i — 1 (it's a simple mathematical substitution). A first idea would be just to iterate each term and calculate it with exponentiation by squaring. However, it takes in the worst case the same complexity as iterating all plate. We need to find something smarter.

2 ^ i + 2 ^ (i + n) + 2 ^ (i + 2 * n) + … + 2 ^ (i + (k – 1) * n) =

= 2 ^ i * 1 + 2 ^ i * 2 ^ n + 2 ^ i * 2 ^ (2 * n) + … + 2 ^ i * 2 ^ ((k – 1) * N) =

= 2 ^ i * (2 ^ 0 + 2 ^ n + 2 ^ (2 * n) + … + 2 ^ ((k – 1) * n)

We reduced the problem to calculate sum S = 2 ^ 0 + 2 ^ n + 2 ^ (2 * n) + … + 2 ^ (X * n).

What’s the value of 2 ^ n * S ? It is 2 ^ n + 2 ^ (2 * n) + 2 ^ (3 * n) + … + 2 ^ ((X + 1) * n). And what you get by making 2 ^ n * S – S ?

2 ^ n * S – S = 2 ^ ((X + 1) * n) – 1

S * (2 ^ n – 1) = 2 ^ ((X + 1) * n) – 1

S = (2 ^ ((X + 1) * n) – 1) / (2 ^ n – 1).

We can calculate both 2 ^ i and S with exponentiation by squaring and the problem is done. For "/" operator, we can use multiplicative inverse (you can read about that and about Fermat Little's theorem, taking care that 10^9 + 7 is a prime number). The time complexity is O(N * logK). Note: that kind of reduction of powers is called “power series” in math.

Alternative solution: For this alternative solution, we don't need to use any special properties of 5. In fact, we can replace 5 by any integer p and still have the same solution. So for now, I shall write p in place of 5.

This suggests a dynamic programming solution: denote dp(x,y) be the number of ways of deleting some digits in the first x digits to form a number that has remainder y (modulo p). For simplicity, we accept “empty” plate be a number that is divisible by p. Writing the DP formula is not difficult. We start with dp(0,0) = 1, and suppose we already have the value dp(x,y). We shall use dp(x,y) to update for dp(x + 1,*), which has two possible cases: either keeping the (x + 1)-th digit or by deleting it. I won't go into much detail here. The answer is therefore dp(N,0).

Clearly, applying this DP directly would time out. For a better algorithm, we resort on the periodicity of the actual plate. The key idea is that, we imagine each digit in the plate as a linear transformation from (x0, x1, .., x(p – 1)) to (y0, y1, y(p-1)). Obviously, (x0, x1, .., x(p — 1)) corresponds to some dp(i, 0), dp(i, 1) .. dp(i, p — 1) and (y0, y1, y(p-1)) corresponds to some (dp(i + 1, 0)), dp((i + 1), 1), ..., dp(i + 1, p — 1) .So we can write X * M(d) = Y, where X and Y are vectors of length p, and M(d) is the matrix of size p * p representing digit d (note that M(d) is independent from X and Y). By multiplying all |a|.K such matrices together, we obtain a transformation from (1, 0, 0, .., 0) to (T0, T1, .., T(p – 1)) where T0 is actually our answer (including the empty plate).

What's the difference? We can group the matrices in groups of length |a|, and lift to the exponent K. That leads to an algorithm with time complexity O(p^3(|a| + log K)), which could be risky. To improve, we should go back to our original DP function and observe that it is actually a linear transformation from (1, 0, 0, .., 0) to (R0, R1, …, R(p – 1)), if we restrict ourselves in the first fragment of length |a|. So instead of multiplying |a| matrices together, we can run DP p times with initial conditions (0, 0, .., 0, 1, 0, .., 0) to obtain the matrix transformation. The overall time complexity becomes O(|a| * p^2 + p^3 log K) 

----------------------------------code---------------------------------------------------------------------------------

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<fstream>
#include<map>
#include<ctime>
#include<set>
#include<queue>
#include<cmath>
#include<vector>
#include<bitset>
#include<functional>
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define REP(i,l,r) for((i)=(l);(i)<=(r);++(i))
#define REP2(i,l,r) for((i)=(l);(i)!=(r);++(i))
using namespace std;

typedef long long LL;
typedef double ld;

const int MAX=200000+2;
const int Mod=(int)1e9+7;

int n,P[MAX];
char a[MAX];

LL power(LL a,LL b)
{
 LL k=1;
 for(;b;b/=2,a=a*a%Mod)
  if(b%2==1)
   k=k*a%Mod;
 return k;
}

int main()
{
#ifndef ONLINE_JUDGE
 freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#endif
 int i,k;
 scanf("%s",a+1);
 scanf("%d",&k);
 n=strlen(a+1);
 P[0]=1;
 REP2(i,1,MAX)
  P[i]=P[i-1]*2%Mod;
 int ans=0;
 //P[i-1]*P[n] + 
 LL c=power(P[n],k);
 LL d=power((P[n]-1+Mod)%Mod,Mod-2);
 REP(i,1,n)
  if(a[i]=='0' || a[i]=='5')
  {
   LL t=(LL)P[i-1]*(c-1+Mod)%Mod*d%Mod;
   ans=(ans+t)%Mod;
  }
 cout<<ans<<endl;
 return 0;
}