codeforce465DIV2——D. Fafa and Ancient Alphabet

概率的计算答案给出的这张图很清楚了，然后因为要求取模，a/b%M=a*b^-1%M=a*inv(b,M)%M;

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdio>

using namespace std;
const int maxn=100000+10;
const int MOD=1000000007;
typedef long long LL;
int n,m;
int s1[maxn],s2[maxn],d[maxn];
void gcd(LL a,LL b,LL &d,LL &x,LL &y){
    if(!b){
        d=a;x=1;y=0;
    }else{
        gcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}
LL inv(LL a,LL n){
    LL d,x,y;
    gcd(a,n,d,x,y);
    return d==1?(x+n)%n:-1;
}
LL dp(int a){
    if(a==n+1)return 0;
    if(s1[a]==s2[a]&&s1[a]!=0&&s2[a]!=0)return dp(a+1);
    if(s1[a]>s2[a]&&s1[a]!=0&&s2[a]!=0)return 1;
    if(s1[a]<s2[a]&&s1[a]!=0&&s2[a]!=0)return 0;
    if(s1[a]==0&&s2[a]!=0)
    return (((m-s2[a])%MOD*inv(m,MOD))%MOD+dp(a+1)%MOD*inv(m,MOD))%MOD;
    if(s1[a]!=0&&s2[a]==0)
    return (((s1[a]-1)%MOD*inv(m,MOD))%MOD+(dp(a+1))%MOD*inv(m,MOD))%MOD;
    if(s1[a]==0&&s2[a]==0)
    return (((m-1)%MOD*inv(2*m,MOD))%MOD+(dp(a+1)%MOD*inv(m,MOD)))%MOD;
}


int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&s1[i]);
    for(int i=1;i<=n;i++)scanf("%d",&s2[i]);
    LL P=dp(1);
    cout<<P;
return 0;
}