Codeforces-935d:fafa and Ancient Alphabet (DP)

D. Fafa and Ancient Alphabet time limit per test 2 seconds memory limit per test, megabytes input standard input output Standard output

Ancient Egyptians is known to has used a large set of symbols to write on the walls of the temples. Fafa and Fifa went to one of the temples and found, non-empty words S1 and S2 of equal lengths on the wall of temple WR Itten one below the other. Since This temple are very ancient, some symbols from the words were erased. The symbols in the set has equal probability for being in the position of any erased symbol.

Fifa challenged Fafa to calculate the probability that S1 are lexicographically greater than S2. Can Fafa with this task?

You know that, i. e. There were m distinct characters in Egyptians ' alphabet, in this problem these characters is denote d by integers from 1 to M in alphabet order. A word X is lexicographically greater than a word y of the same length, if the words was same up to some position, and the n the word x has a larger character, than the word y.

We can prove the probability equals to some fraction, where P and Q is coprime integers, and.  Print as the answer the value, i. e. Such a non-negative integer less than 109 + 7, such this, where means that A and B Give the same remainders when divided by M. Input

The first line contains-integers n and M (1≤n, m≤105)-the length of each of the both words and the size of the A Lphabet, respectively.

The second line contains n integers a1, a2, ..., an (0≤ai≤m)-the symbols of S1. If ai = 0, then the symbol at position I was erased.

The third line contains n integers representing S2 with the same format as S1. Output

Print the value, where P and Q are coprime and was the answer to the problem. Examples input Copy

1 2
0
1

Output

500000004

Input Copy

1 2
1
0

Output

0

Input Copy

7
0 9 0 11 1 0
13 15 12 0

Output

230769233

Note

In the first sample, the first word can is converted into (1) or (2). The second option is the one and that'll make it lexicographically larger than the second word. So, the answer to the problem would be, which is 500000004, because.

In the second example, there are no replacement for the zero in the second word that'll make the first one Lexicographica lly larger. So, the answer to the problem are, that's 0.

Idea: DP. D[i][0] Represents the [1,i]s1=s2 (dictionary order) scheme number; d[i][1] represents the number of scenarios for S1>S2. The final answer is the total number of d[n][1]/programs.

#include <bits/stdc++.h> using namespace std;
const int MAX=1E6;
const int mod=1e9+7;
typedef __int64 LL;
int A[max],b[max];
ll D[max][2];
    ll POW (ll x,ll N) {ll res=1;
        while (n) {if (n&1) Res=res*x%mod;
        X=x*x%mod;
    n/=2;
} return res;
    } int main () {ll n,m,tot=0;
    cin>>n>>m;
    for (int i=1;i<=n;i++) scanf ("%d", &a[i]);
    for (int i=1;i<=n;i++) scanf ("%d", &b[i]);
    memset (d,0,sizeof D);
    D[0][0]=1;
        for (int i=1;i<=n;i++) {tot+= (a[i]==0);
        tot+= (b[i]==0);
                if (A[i]&&b[i]) {if (A[i]==b[i]) {d[i][0]=d[i-1][0];
            D[I][1]=D[I-1][1];
                } else if (A[i]>b[i]) {d[i][1]= (d[i-1][1]+d[i-1][0])%mod;
            d[i][0]=0;
                } else {d[i][1]=d[i-1][1];
            d[i][0]=0;
     }
        }   else if (a[i]==0&&b[i]==0) {d[i][0]=d[i-1][0]*m%mod;
        D[i][1]= ((d[i-1][1]* ((m*m)%mod)%mod+d[i-1][0]* ((m-1) *M/2)%mod)%mod)%mod;
            } else if (a[i]==0) {d[i][0]=d[i-1][0]%mod;
        D[i][1]= ((d[i-1][0]* (m-b[i))%mod+ (d[i-1][1]*m)%mod)%mod;
            } else if (b[i]==0) {d[i][0]=d[i-1][0]%mod;
        D[i][1]= ((d[i-1][0]* (a[i]-1))%mod+ (d[i-1][1]*m)%mod)%mod;
    }} cout<< (D[n][1]*pow (POW (M,tot), MOD-2))%mod<<endl;
return 0;
 }